Need help

Ok im using structures in a programme i got from university

I was wondering how I i can use the information entered in one void function by the user in a new void function

im not that good at explaining because I havn't a clue what im doin but ill post up an example (not full programme and doesnt work)

I want to use the value entered for char student_no to be used in my IF statment in the void function below

Sorry if you cant understand what im looking for
Thanks :)

Code:

---


void borrow_a_book (LIBRARY books_array[])
{
    int book_code;
    int dept;
    char student_no[9];
    char exh1[20];
   
    /*Get the books library code*/
    printf("\n\nEnter the book's library code that you wish to borrow (1 digit, except 0) : ");
    do
    scanf("%d", &book_code);
    while (book_code <= 0);
       
   
       
    /*Find the department for the book is placed in*/
    dept = search_database (books_array, book_code);
   
    /*Scan through the database to see if it is available*/
    if (dept == MAX_BOOKS)
        printf("\n\Sorry this book is already rented\n");
    else
    {
        printf("\nPlease enter your student number (9) :");    /*Student number entered after the if statement*/
        scanf("%d", &student_no);
        printf("\n\nBook has has now been borrowed",book_code);
        printf("\nPress enter to continue");
        gets(exh1);
        gets(books_array[dept].name);
    }
}


void return_a_book (LIBRARY books_array[])
{
    int book_code;
    int dept;
    char student_no[9];
    char student_no2[9];
   
    /*Book is marked as returned to database when it is available*/
   
    /*First we find the book code*/
    printf("\n\nBook code to find (2 digits, except 0) : ");
    do
    scanf("%d", &book_code);
    while (book_code <= 0);
       
    /*Find the department the book is in*/
    dept = search_database (books_array, book_code);
   
    /*Scan through database to see if book has been returned*/
    printf("Please enter your student number    : ");    /*student number entered before IF statement so programme knows who has the book*/
    scanf("%d", &student_no2);
    if (student_no2 != student_no)
    {
        printf("\n\n Sorry you are not the student who borrowed this book");
    }
    else
        if (dept != MAX_BOOKS)
        {
            printf("\n\n The book has already been returned");
        }
        else
            {
                printf("\n\nThe book has now been returned, thank you");
                books_array[dept].code = 0;
            }
}

---

Make it a global variable.

Dont really know what a global variable is :(

I did put it under my book structure... but it still wont work

when i run the whole programme even when the student numbers are equal it still prints to screen "sorry you are not the student with this book"

Code:

---


#include<stdio.h>
#include<conio.h>
#include<string.h>
#define MAX_BOOKS 5

/*structure definitions*/
struct date
{
    int day;
    int month;
    int year;
};

typedef struct date DATE;
   
struct books
{
    int code;              /*This is the libary code at which the book is stored in the system*/
    char name[20];          /*Title of book*/
    int isbn[13] ;          /*isbn of book*/
    char surname[26];      /*Authors surname*/
    char first_name[11];    /*Authors First name*/
    char cust[20];          /*Name of customer*/
    char student_no[9];    /*student number*/
    char student_no2[9];    /*student number to compare to*/
    int dept;
    DATE recieved;
};
typedef struct books LIBRARY;
   
/*Function prototypes*/
void add_a_book (LIBRARY[]);
void remove_a_book (LIBRARY[]);
void borrow_a_book(LIBRARY[]);
void return_a_book(LIBRARY[]);
void search_by_book_name(LIBRARY[]);
void display_a_book(LIBRARY[]);
void display_book_details(LIBRARY *);
void display_book_details2(LIBRARY *);
void init_database(LIBRARY[]);
void book_list(LIBRARY[]);
int search_database(LIBRARY[], int);
int search_database2(LIBRARY[], char[]);
int menu(void);

---

Quote:

---

Originally Posted by Soulsurvivor

I want to use the value entered for char student_no to be used in my IF statment in the void function below

---

Do not say "void function". Name the function that you are talking about. If you want to say that the function returns void, say that, not "void function".

Quote:

---

Originally Posted by Soulsurvivor

Dont really know what a global variable is :(

I did put it under my book structure... but it still wont work

when i run the whole programme even when the student numbers are equal it still prints to screen "sorry you are not the student with this book"

Code:

---


#include<stdio.h>
#include<conio.h>
#include<string.h>
#define MAX_BOOKS 5

/*structure definitions*/
struct date
{
    int day;
    int month;
    int year;
};

typedef struct date DATE;
   
struct books
{
    int code;              /*This is the libary code at which the book is stored in the system*/
    char name[20];          /*Title of book*/
    int isbn[13] ;          /*isbn of book*/
    char surname[26];      /*Authors surname*/
    char first_name[11];    /*Authors First name*/
    char cust[20];          /*Name of customer*/
    char student_no[9];    /*student number*/
    char student_no2[9];    /*student number to compare to*/
    int dept;
    DATE recieved;
};
typedef struct books LIBRARY;
   
/*Function prototypes*/
void add_a_book (LIBRARY[]);
void remove_a_book (LIBRARY[]);
void borrow_a_book(LIBRARY[]);
void return_a_book(LIBRARY[]);
void search_by_book_name(LIBRARY[]);
void display_a_book(LIBRARY[]);
void display_book_details(LIBRARY *);
void display_book_details2(LIBRARY *);
void init_database(LIBRARY[]);
void book_list(LIBRARY[]);
int search_database(LIBRARY[], int);
int search_database2(LIBRARY[], char[]);
int menu(void);

---

---

A global variable is a variable that has global scope, i.e. is declared outside your functions and can be accessed in any of them.

Just type:

Code:

---

    char student_no[9];    /*student number*/
    char student_no2[9];    /*student number to compare to*/

---

above your function prototyping and remove the same lines from the functions you were referring to.

Now, everytime you say student_no[0] in your functions it will refer to the same thing. Is this what you were trying to do?

I have my reservations about claudiu's suggestion because global variables generally should be avoided. They tend to make it more difficult to understand, debug and maintain a program.

Quote:

---

Originally Posted by laserlight

Do not say "void function". Name the function that you are talking about. If you want to say that the function returns void, say that, not "void function".

---

Sorry i don't really know the terms.. i just learn from example, ive only been programming for 3 months so im pretty bad :(

So any idea whats wrong?

Quote:

---

Originally Posted by claudiu

A global variable is a variable that has global scope, i.e. is declared outside your functions and can be accessed in any of them.

---

That is kind of misleading. I would almost say that is incorrect. But whatever. Here is what I get..

Code:

---

#include <stdio.h>

int x = 3;

int main(void)
{
  printf("The value of x is: %d\n", x);

  {
    int x = 25;
    printf("The value of x is now : %d\n", x);
  }

  printf("The value of x again is: %d\n", x);

  return 0;
}

---

[cd@localhost oakland]$ gcc -Wall -Wextra -ansi -pedantic moron.c -o moron
[cd@localhost oakland]$ ./moron
The value of x is: 3
The value of x is now : 25
The value of x again is: 3
[cd@localhost oakland]$

Quote:

---

printf("The value of x again is: %d\n", x);

---

It's more accurate to say "The value of x is still: %d\n", because the value never actually changes. You are never changing the value of X. You are changing which X you're talking about.


Quzah.

Quote:

---

Originally Posted by quzah

It's more accurate to say "The value of x is still: %d\n", because the value never actually changes. You are never changing the value of X. You are changing which X you're talking about.


Quzah.

---

That is incorrect. The identifier x never changes. What changes is the value because I'm rebinding x

The value of x never changes in your program. THE x changes, not its value. Want proof? Make them const, then tell me you're changing the value of x. (And yes, we've all seen the pointer trick to change constants, so don't bother looking it up.) You aren't changing the value of anything. You're changing which object you're talking about.

Edit: If you still don't believe it, turn on name mangling and dump the output and make yourself a humble pie.


Quzah.

Quote:

---

Originally Posted by quzah

The value of x never changes in your program. THE x changes, not its value. Want proof? Make them const, then tell me you're changing the value of x. (And yes, we've all seen the pointer trick to change constants, so don't bother looking it up.) You aren't changing the value of anything. You're changing which object you're talking about.


Quzah.

---

No, x is an identifier. It names an object. Yes, the object might change, but the identifier x itself doesn't change.

So what's your point? You still aren't changing the value of X.


Quzah.

Overworked_PhD:

Do you like being a jackass?

Quote:

---

Originally Posted by Overworked_PhD

TROLL TROLL TROLL

---

Hahah. I was actually hoping you could produce something intelligent for a change. I could have sworn you posted something.

Oh nope. Nevermind! Same aggressive attitude towards all the other posters in a futile attempt to raise your comments to some adequate standard of intelligence and to "show'em" who knows C best.

Quote:

---

Originally Posted by quzah

So what's your point? You still aren't changing the value of X.


Quzah.

---

Did I say I was changing the value of x? I thought I said I was rebinding x and not changing the value (via assignment).

Quote:

---

Originally Posted by Overworked_PhD

Did I say I was changing the value of x?

---

Yes. Yes you did:

Quote:

---

Code:

---

int main(void)
{
  printf("The value of x is: %d\n", x);

  {
    int x = 25;
    printf("The value of x is now : %d\n", x);
  }

  printf("The value of x again is: %d\n", x);

---

---

You imply that the value of X changes. Imply? Did I just say IMPLY? No, you SAY it changes. It doesn't. I pointed it out, you said it did change:

Quote:

---

Originally Posted by Overworked_PhD

That is incorrect. The identifier x never changes. What changes is the value because I'm rebinding x

---

Be less dumb.


Quzah.

Quote:

---

Originally Posted by quzah

Yes. Yes you did:You imply that the value of X changes. Imply? Did I just say IMPLY? No, you SAY it changes. It doesn't. I pointed it out, you said it did change:Be less dumb.


Quzah.

---

Yes, I said x changes because I 'rebinded' x. For whatever reasons, you seem to be under the impression that x changed because I reassigned it. Reassignment and rebinding are two different beast.

Do you want me to recap the whole thing, because you're too much of an idiot to admit you made an error and were corrected on it?

You: The value of x changes.
Me: No, the values never change. The x changes.
You: The value changes because I rebinded it!
Me: No, the values never change. The x changes.
You: Nu uh! I never said the value changes!
Me: Yes you did, right there, and there and there.
You: Nu uh!

You didn't say "x changes". You said "what changes is the value". No it doesn't. The values of x never change. The object changes. No values change at all.

I didn't flat out call you an idiot from the start, I said "it's more accurate to say..." and you got all butthurt over it. Stop being such a baby or get the ........ off the internet.


Quzah.

Quote:

---

Originally Posted by quzah

Do you want me to recap the whole thing, because you're too much of an idiot to admit you made an error and were corrected on it?

You: The value of x changes.
Me: No, the values never change. The x changes.
You: The value changes because I rebinded it!
Me: No, the values never change. The x changes.
You: Nu uh! I never said the value changes!
Me: Yes you did, right there, and there and there.
You: Nu uh!

You didn't say "x changes". You said "what changes is the value". No it doesn't. The values of x never change. The object changes. No values change at all.

I didn't flat out call you an idiot from the start, I said "it's more accurate to say..." and you got all butthurt over it. Stop being such a baby or get the ........ off the internet.


Quzah.

---

Quit twisting words. You thought that x changed. I pointed out that the object and not x itself changed.

Quote:

---

Originally Posted by Overworked_PhD

Quit twisting words. You thought that x changed. I pointed out that the object and not x itself changed.

---

No, I'm quite clear on how the scope of objects work with regards to reusing the same name. That's why I said clearly:

Quote:

---

Originally Posted by quzah

It's more accurate to say "The value of x is still: %d\n", because the value never actually changes. You are never changing the value of X. You are changing which X you're talking about.

---

See? There's no confusion on my part. There's not even a hint of confusion. I have never twisted any words in any post here, I have been implicitly clear as to exactly what I have been talking about. Since you're obviously still confused, I'll say it one last time:

The values of x (both of them) never change. There are two distinct objects named x, and you reference each one separately. Their values never change. You said they do, you were wrong.


Quzah.

Quote:

---

Originally Posted by quzah

No, I'm quite clear on how the scope of objects work with regards to reusing the same name. That's why I said clearly:See? There's no confusion on my part. There's not even a hint of confusion. I have never twisted any words in any post here, I have been implicitly clear as to exactly what I have been talking about. Since you're obviously still confused, I'll say it one last time:

The values of x (both of them) never change. There are two distinct objects named x, and you reference each one separately. Their values never change. You said they do, you were wrong.


Quzah.

---

Saying 'You are changing which X you're talking about.' is a really really sloppy way of saying that you are changing the object. I think you made this statement because you still aren't clear about the relationship between identifiers and objects. So yes, there is a hint of confusion on your part since you failed to recognize the difference between an identifier and an object.

Also

Quote:

---

The values of x (both of them) never change.

---

There aren't two values of x. There are two objects that have two different values.

Quote:

---

Originally Posted by Overworked_PhD

Also



There aren't two values of x. There are two objects that have two different values.

---

LOL. Are you serios dude? This reads, there aren't two values, there are two values.

Quote:

---

Originally Posted by Overworked_PhD

Saying 'You are changing which X you're talking about.' is a really really sloppy way of saying that you are changing the object. I think you made this statement because you still aren't clear about the relationship between identifiers and objects. So yes, there is a hint of confusion on your part since you failed to recognize the difference between an identifier and an object.

---

You're an idiot who is trying to talk his way out of looking bad. Seriously dude, "saving face" isn't that important. Besides, everyone here already knows you are an idiot. So give it up already.

There was no "really sloppy way" in my first post, because I implicitly said what I meant. Want it quoted again, because you can't figure out what the left arrow on the forum does? Ok tyke, here we go again:

Quote:

---

Quote:

---

because the value never actually changes. You are never changing the value of X. You are changing which X you're talking about.


Quzah.

---

That is incorrect. The identifier x never changes. What changes is the value because I'm rebinding x

---

No, dumbass. "the value" doesn't change. You never change any values. You change the object you're talking about. Period. Now go away and come back when you're not a complete raving idiot.

If you can't stand the fire, get off the stove, idiot.


Quzah.

Quote:

---

Originally Posted by quzah

You're an idiot who is trying to talk his way out of looking bad. Seriously dude, "saving face" isn't that important. Besides, everyone here already knows you are an idiot. So give it up already.

There was no "really sloppy way" in my first post, because I implicitly said what I meant. Want it quoted again, because you can't figure out what the left arrow on the forum does? Ok tyke, here we go again:No, dumbass. "the value" doesn't change. You never change any values. You change the object you're talking about. Period. Now go away and come back when you're not a complete raving idiot.

If you can't stand the fire, get off the stove, idiot.


Quzah.

---

No. If I do something like the following...

Code:

---

int x = 0;
x = 3

---

The value of the object is zero when I initialize it. Now when I reassign the value 3 to the same object, it becomes three. In this case, the value changes, but the underlying object itself doesn't.

Quote:

---

Originally Posted by claudiu

LOL. Are you serios dude? This reads, there aren't two values, there are two values.

---

I'm drawing a distinction between the identifier and the object itself.

Quote:

---

not the underlying object itself doesn't.

---

O_o

I'm now in desperate need of ibuprofen.

Soma

Quote:

---

Originally Posted by Overworked_PhD

No. If I do something like the following...

---

IF you had actually done the following, THEN you would have been correct in saying that the value has changed. You didn't. So none of this actually matters or supports your argument in any way.

Quote:

---

Originally Posted by Overworked_PhD

Code:

---

int x = 0;
x = 3

---

The value of the object is zero when I initialize it. Now when I reassign the value 3 to the same object, it becomes three. In this case, the value changes, but not the underlying object itself doesn't.

---

Congratulations! You understand how to change a value. That has nothing to do with the post I'm talking about, except for the fact that that isn't what you did. You said you did. You didn't. I pointed it out. You tried to say you were right. You weren't. End of story.

All that's left is you trying to provide some wild circular-logic and enough smoke screens to confuse people who don't really know that you were wrong in the first place. (Hint: Not me.) Your wording was wrong, as I pointed out, and ever since then, you've been trying to prove yourself right. You aren't. You may understand that in fact the value did not change, and I really don't care. That's not what I was saying. I didn't say "you don't understand the concept". I said you worded it incorrectly. Ever since, all you've done is try to talk your way out of a hole, by saying you said this or that. You didn't, and you just look stupid.

Quzah.

Quote:

---

Originally Posted by quzah

IF you had actually done the following, THEN you would have been correct in saying that the value has changed. You didn't. So none of this actually matters or supports your argument in any way.Congratulations! You understand how to change a value. That has nothing to do with the post I'm talking about, except for the fact that that isn't what you did. You said you did. You didn't. I pointed it out. You tried to say you were right. You weren't. End of story.

All that's left is you trying to provide some wild circular-logic and enough smoke screens to confuse people who don't really know that you were wrong in the first place. (Hint: Not me.) Your wording was wrong, as I pointed out, and ever since then, you've been trying to prove yourself right. You aren't. You may understand that in fact the value did not change, and I really don't care. That's not what I was saying. I didn't say "you don't understand the concept". I said you worded it incorrectly. Ever since, all you've done is try to talk your way out of a hole, by saying you said this or that. You didn't, and you just look stupid.

Quzah.

---

I'm changing the value because I was doing a re-assignment. If I would do this same thing inside a pair of braces in main(), this would be doing a rebind because I'm actually changing the object and not the identifier. Re-read your posts. You always asserted that I was changing 'x'. I pointed out that I was changing the object, and not x itself.

Its like talking to a wall.

I think he just likes to argue. There is no winning.

Quote:

---

Originally Posted by prog-bman

Its like talking to a wall.

I think he just likes to argue. There is no winning.

---

It's not that. He probably just sits in a cubicle all day with no friends to talk to so he goes online and trolls forums, in desperate need for human contact.

It's just that now, it is obvious why he doesn't have friends.

Quote:

---

Originally Posted by Overworked_PhD

I'm changing the value because I was doing a re-assignment.

---

No you aren't! What you're trying to say is the same as this:

Code:

---

void fun( void )
{
    int x = 10;
    printf( "The value of x changed! %d\n", x );
}
int main( void )
{
    int x = 5;
    printf("The value of x is %d!\n", x );
    fun( );
    printf("The value of x is changed back to %d!\n", x );
    return 0;
}

---

No it didn't! X's value didn't change! You changed Xs, not X's value. Notice the distinction between the apostrophe and the lack there of. This is very much a language issue, not a value issue.

Quote:

---

You always asserted that I was changing 'x'. I pointed out that I was changing the object, and not x itself.

---

You changed which object you were talking about, not the value of the object. The value of the first object x never changes. Hell, the value of the second object x never changes. You aren't changing the value of x. You are changing which x you're talking about. That's why I said you worded it incorrectly.


Quzah.

Quote:

---

Originally Posted by quzah

No you aren't! What you're trying to say is the same as this:

Code:

---

void fun( void )
{
    int x = 10;
    printf( "The value of x changed! %d\n", x );
}
int main( void )
{
    int x = 5;
    printf("The value of x is %d!\n", x );
    fun( );
    printf("The value of x is changed back to %d!\n", x );
    return 0;
}

---

No it didn't! X's value didn't change! You changed Xs, not X's value. Notice the distinction between the apostrophe and the lack there of. This is very much a language issue, not a value issue.You changed which object you were talking about, not the value of the object. The value of the first object x never changes. Hell, the value of the second object x never changes. You aren't changing the value of x. You are changing which x you're talking about. That's why I said you worded it incorrectly.

Quzah.

---

No, again, you changed the object and not the identifier 'x' itself in fun(). Yes, the value of the first object named 'x' never changes. And yes, the value of the second object named 'x' never changes. What is different between 'x' in fun() and 'x' in main() is what 'x' refers to. In func(), 'x' refers to an object. In main(), 'x' refers to a different object. In both cases, what changed was the object and not the identifier 'x' itself.

How about this.

Code:

---

#include <stdio.h>

int x = 3;

int main(void)
{
  printf("The value of x is: %d\nAddress of x is: %p\n", x, &x);

  {
    int x = 25;
    printf("The value of x is now : %d\nAddress of x is now: %p\n", x, &x);
  }

  printf("The value of x again is: %d\nAddress of x again is: %p\n", x, &x);

  return 0;
}

---

It appears that these are two variables with the same name.

Quote:

---

Originally Posted by Subsonics

How about this.

Code:

---

#include <stdio.h>

int x = 3;

int main(void)
{
  printf("The value of x is: %d\nAddress of x is: %p\n", x, &x);

  {
    int x = 25;
    printf("The value of x is now : %d\nAddress of x is now: %p\n", x, &x);
  }

  printf("The value of x again is: %d\nAddress of x again is: %p\n", x, &x);

  return 0;
}

---

It appears that these are two variables with the same name.

---

What is the same is the identifier 'x'.

Now going slightly off topic, the ANSI/ISO C standards seems to imply that a variable is a named object. However, I can't find anything that actually supports what they are implying.

Quote:

---

Originally Posted by Overworked_PhD

No, again, you changed the object and not the identifier 'x' itself in fun().

---

Of course I did. That's what you did:

Code:

---

int x; <-- make an object named x
{
    int x; <-- make a second object called x
    {
        int x; <-- make a third object named x

---

WTF do you think it does if it's not making a new object? Of course it's making a new object! That's the entire point! You aren't changing values as you said you were. You're changing objects.

At this point, this conversation leads me to believe that any conversations you have with "dropped names" ends in them calling you an idiot.


Quzah.

Quote:

---

Originally Posted by Overworked_PhD

That is incorrect. The identifier x never changes. What changes is the value because I'm rebinding x

---

Read your last post, and then read this. These are your own words.

Also, better stick to fluid mechanics. You may actually have a future there.

Quote:

---

Originally Posted by Subsonics

It appears that these are two variables with the same name.

---

That's the whole point. He's not changing the value of x. He's making a new variable. He either has completely no clue as to what variable scope is, or he's as always, trolling. Or maybe he's just really really stupid.


Quzah.

Quote:

---

Originally Posted by quzah

Of course I did. That's what you did:

Code:

---

int x; <-- make an object named x
{
    int x; <-- make a second object called x
    {
        int x; <-- make a third object named x

---

WTF do you think it does if it's not making a new object? Of course it's making a new object! That's the entire point! You aren't changing values as you said you were. You're changing objects.

At this point, this conversation leads me to believe that any conversations you have with "dropped names" ends in them calling you an idiot.


Quzah.

---

No, you just said, and previously said

Code:

---

No it didn't! X's value didn't change! You changed Xs, not X's value.

---

In other words, just like in all your other posts, you were talking about the identifier and not the object.

Changed Xs implies changing objects.

What English language do you speak?

Quote:

---

Originally Posted by claudiu

Read your last post, and then read this. These are your own words.

Also, better stick to fluid mechanics. You may actually have a future there.

---

What was the value before 25?

Quote:

---

Originally Posted by prog-bman

Changed Xs implies changing objects.

What English language do you speak?

---

You don't change 'x'. 'x' in fun() is the same as 'x' main(). If 'x' really did change, then in fun(), you would have something 'x_1', and in main(), you would something have 'x_2', However you don't, because 'x' is 'x'.

Quote:

---

Originally Posted by Overworked_PhD

What was the value before 25?

---

Nothing. The inner variable x didn't exist before it was 25. Case and point, dumbass:

Code:

---

#include<stdio.h>
int main( void )
{
    int x = 10;
    int *y = &x;
    {
        int x = 20;
        printf( "x, outer scope, is %d, inner scope is %d\nWhat a dumbass!\n", *y, x );
    }
    return 0;
}

---

One last time, because it really can't be stated enough: "What a dumbass!"


Quzah.

I have two batteries in my mouse and two batteries on my desk.

I changed to the batteries on my desk.

I still have batteries. But they aren't the same batteries, so I CHANGED batteries.

Quote:

---

Originally Posted by quzah

Nothing. The inner variable x didn't exist before it was 25. Case and point, dumbass:

Code:

---

#include<stdio.h>
int main( void )
{
    int x = 10;
    int *y = &x;
    {
        int x = 20;
        printf( "x, outer scope, is %d, inner scope is %d\nWhat a dumbass!\n", *y, x );
    }
    return 0;
}

---

One last time, because it really can't be stated enough: "What a dumbass!"


Quzah.

---

The prior value was garbage. I actually changed the value when I assigned 25 to object. And now, how does your code related any thing that we are talking about? I don't ever recall mentioning pointers. Do you?

Oh wait, I see. You want to confuse us by introducing aliases!

Quote:

---

Originally Posted by Overworked_PhD

The prior value was garbage. I actually changed the value when I assigned 25 to object. And now, how does your code related any thing that we are talking about? I don't ever recall mentioning pointers. Do you?

---

It wasn't "garbage", it didn't exist. Variables don't exist before you declare them. Physical memory exists some place, but the variable, the object, does not. Therefore, there is no such value. There is nothing.

What exactly do you think that pointer is doing there? Do you think non-dynamic variables somehow magically vanish when a new scope block is entered? Come on smart guy, go back to your usenet and ask them what that pointer is doing there. Then come back and tell me how many of them said, "What a dumbass!" -- or did they all killfile you already?


Quzah.

Quote:

---

Originally Posted by quzah

It wasn't "garbage", it didn't exist. Variables don't exist before you declare them. Physical memory exists some place, but the variable, the object, does not. Therefore, there is no such value. There is nothing.

What exactly do you think that pointer is doing there? Do you think non-dynamic variables somehow magically vanish when a new scope block is entered? Come on smart guy, go back to your usenet and ask them what that pointer is doing there. Then come back and tell me how many of them said, "What a dumbass!" -- or did they all killfile you already?


Quzah.

---

I actually *declared* the variables when I did

Code:

---

{
  int x = 25; /*<--I declare and define the variable here
}

---

Actually, this two steps combined into one. First I declare a variable. This variable has some garbage value. The I assigned it a value of 25. In other words, the above code snippet would be equivalent to

Code:

---

{
  int x;  /*This comes into existence and has some garbage value*/
  x = 25
}

---

Correct but this wouldn't work as quzah was saying.

Because x isn't declared yet, so it doesn't exist to the program until it comes into scope.

Code:

---

int main()
{
    printf("%d", x);
    {
        int x = 25;
        printf("%d", x);
    }
}

---

Quote:

---

Originally Posted by prog-bman

Correct but this wouldn't work as quzah was saying.

Because x isn't declared yet, so it doesn't exist to the program until it comes into scope.

Code:

---

int main()
{
    printf("%d", x);
    {
        int x = 25;
        printf("%d", x);
    }
}

---

---

Okay, gotcha. Off topic though, the above code would work in some functional programming languages like Haskell.

Quote:

---

Originally Posted by Overworked_PhD

Okay, gotcha. Off topic though, the above code would work in some functional programming languages like Haskell.

---

Stop creating SMOKE SCREENS! Nobody here cares about Haskell!

Be a man!

Quote:

---

Originally Posted by Overworked_PhD

Okay, gotcha. Off topic though, the above code would work in some functional programming languages like Haskell.

---

Oddly enough, they don't have a Haskell sub-forum on cprogramming.com

The person that gave the following response was (is) the Director of the Super Computing Labs at UC San Diego

Long winded question about objects and identifiers - comp.lang.c | Google Groups

Quote:

---

> So what gets changed? The (value of the) object (named 'x')? Or does
> 'x' itself change? Also, would 'x' be considered an identifier?

"You have two distinct objects, both named ``x''. The first, which we
can call the outer x, is created by the definition ``int x = 3;''.
It has static storage duration, meaning that it exists for the
entire lifetime of your executing program. The second, the inner
x, is created by the definition ``int x = 25;''. It has automatic
storage duration, meaning that it exists only during the execution
of the block (compound statement) that contains its definition.

When the first and third printf calls are executed, only the outer
x exists. When the second printf call is executed, both objects
exist simultaneously, but only the inner one is visible, because
it hides the outer one.

The only thing these two objects have in common is that they both
have the same name and the same type. You could change the name
of the inner x to y, and the program would do exactly the same thing.

Yes, x is an identifier.

This kind of hiding is usually not a good idea, because it can cause
confusion. "

---

Quote:

---

Originally Posted by Overworked_PhD

The only thing these two objects have in common is that they both
have the same name and the same type.

---

Yes. This means that you are NOT CHANGING THE VALUE of x. Stop running to other people to wipe your ass when you say stupid things.

Um, I see no conflicting views here from what have already been discussed in this thread, those are two different variables.

Quote:

---

Originally Posted by claudiu

Yes. This means that you are NOT CHANGING THE VALUE of x. Stop running to other people to wipe your ass when you say stupid things.

---

'x' does get changed. This is because something like

Code:

---

{
 int x = 25;
}

---

is the same as

Code:

---

{
 int x; /*x initially has a garbage value*/
  x = 25;
}

---

Ie, the value of x changes from a 'junk value' to 25. I'm now going to look up corresponding passages in ANSI/ISO C that supports this fact since you will debate this point until the bitter end. I believe technically x has a tentative definition, whose value is 'junk' until it completes the translation unit. Then, after the completion of the translation unit, the value of x becomes (changes to) 25.

Quote:

---

I believe technically x has a tentative definition, whose value is 'junk' until it completes the translation unit.

---

^_^

This thread keeps getting better!

Soma

Quote:

---

Originally Posted by Overworked_PhD

'x' does get changed. This is because something like

Code:

---

{
 int x = 25;
}

---

is the same as

Code:

---

{
 int x; /*x initially has a garbage value*/
  x = 25;
}

---

Ie, the value of x changes from a 'junk value' to 25. I'm now going to look up corresponding passages in ANSI/ISO C that supports this fact since you will debate this point until the bitter end. I believe technically x has a tentative definition, whose value is 'junk' until it completes the translation unit. Then, after the completion of the translation unit, the value of x becomes (changes to) 25.

---

You are debating a whole different point now then you were the day before. You know exactly what I meant in the previous message, yet you continue to act surprised.

You were previously arguing that the value of 'x' changes from 3 to 25 in your original example and then back to 3, and now have pulled back to arguing what we were saying in the first place.

Also, you don't have a PhD. Anyone with half a brain could realize that. You just enjoy being a fraud, and that is so pathetic it almost makes me cry.

Quote:

---

Originally Posted by claudiu

You are debating a whole different point now then you were the day before. You know exactly what I meant in the previous message, yet you continue to act surprised.

---

No i don't. Maybe this is due to the fact that you tend to be ambiguous in your responses.

Quote:

---

Originally Posted by Overworked_PhD

The person that gave the following response was (is) the Director of the Super Computing Labs at UC San Diego

---

Did you snip out the part where he ended his note with "You're a dumbass!"

Because really, that's what he's telling you. You're a dumbass.

Quote:

---

"You have two distinct objects, both named ``x''. The first, which we
can call the outer x, is created by the definition ``int x = 3;''.
It has static storage duration, meaning that it exists for the
entire lifetime of your executing program. The second, the inner
x, is created by the definition ``int x = 25;''. It has automatic
storage duration, meaning that it exists only during the execution
of the block (compound statement) that contains its definition.

---

Yeah, I already said that. Thanks for telling me what I already know, and what I already told you.

Quote:

---

When the first and third printf calls are executed, only the outer
x exists. When the second printf call is executed, both objects
exist simultaneously, but only the inner one is visible, because
it hides the outer one.

---

I told you this also.

Quote:

---

The only thing these two objects have in common is that they both
have the same name and the same type. You could change the name
of the inner x to y, and the program would do exactly the same thing.

Yes, x is an identifier.

This kind of hiding is usually not a good idea, because it can cause
confusion. "

---

Translation: "You are not changing the value of x, you are using a different x." As I already told you.

Translation: "You're a dumbass!"

Thanks for telling me what I already knew, and what I already told you. One more time, all together now:

"You're a dumbass."


Quzah.

Quote:

---

Originally Posted by quzah

"You're a dumbass!


Quzah.

---

Amen.

Quote:

---

Originally Posted by quzah

Did you snip out the part where he ended his note with "You're a dumbass!"

Because really, that's what he's telling you. You're a dumbass.Yeah, I already said that. Thanks for telling me what I already know, and what I already told you.I told you this also.Translation: "You are not changing the value of x, you are using a different x." As I already told you.

Translation: "You're a dumbass!"

Thanks for telling me what I already knew, and what I already told you. One more time, all together now:

"You're a dumbass."


Quzah.

---

Keith speaks in proper technical terms. You don't. Your responses are just plain sloppy. If anyone has to questions your response, like what I did, then it means you suck at technical responses.

Quote:

---

Originally Posted by Overworked_PhD

Keith speaks in proper technical terms. You don't. Your responses are just plain sloppy. If anyone has to questions your response, like what I did, then it means you suck at technical responses.

---

Whaaa whaaa whaaa. Call a whaaambulance!

or

I'm tellling moooom!

Quote:

---

Originally Posted by claudiu

Whaaa whaaa whaaa. Call a whaaambulance!

or

I'm tellling moooom!

---

My current employer, Kodak, won't even give potential programmers a second interview if the person they are interviewing gives sloppy technical responses. Here is what my manager told me. "People that give sloppy technical responses demonstrate that they don't care enough to pay attention to the details. We don't need people like that working here."

Actually I know a few major tech companies in the San Francisco Bay Area that hold similar views.

Quote:

---

Originally Posted by Overworked_PhD

Keith speaks in proper technical terms. You don't.

---

I don't need to speak in technical terms. If you can't figure out that "you are changing which X you are talking about" means "you have two different objects both named X", then ... and I'll be technical here so you understand it:

"You're a dumbass!"


Quzah.

Quote:

---

Originally Posted by quzah

I don't need to speak in technical terms. If you can't figure out that "you are changing which X you are talking about" means "you have two different objects both named X", then ... and I'll be technical here so you understand it:

"You're a dumbass!"


Quzah.

---

No, it means your response is ambiguous. I also know that the College of Electrical Engineering and Computer Science at UC Berkeley would fail you for giving such a poor response.

Quote:

---

Originally Posted by Overworked_PhD

My current employer, Kodak, won't even give potential programmers a second interview if the person they are interviewing gives sloppy technical responses. Here is what my manager told me. "People that give sloppy technical responses demonstrate that they don't care enough to pay attention to the details. We don't need people like that working here."

Actually I know a few major tech companies in the San Francisco Bay Area that hold similar views.

---

If your current employer Kodak prefers someone with wrong technical responses over someone with sloppy technical responses that is entirely his problem. However, I have a feeling your employer has no idea about the nonsensical rubbish you troll about on forums all the time, otherwise you would be living on your unemployment benefits right now.

Quote:

---

Originally Posted by Overworked_PhD

No, it means your response is ambiguous. I also know that the College of Electrical Engineering and Computer Science at UC Berkeley would fail you for giving such a poor response.

---

Yeah right, like you ever graduated anything.

Quote:

---

Originally Posted by claudiu

Yeah right, like you ever graduated anything.

---

I actually hold a B.S in Mechanical Engineering from the University of Wisconsin with a minor in Modern Physics.

No one but you cares, Chad. The fact remains, you are wrong, and I am not. Thanks for playing.


Quzah.

Quote:

---

Originally Posted by Overworked_PhD

I actually hold a B.S in Mechanical Engineering from the University of Wisconsin with a minor in Modern Physics.

---

Congratulations! You should stick to that then. You have given us a wonderful link to your google group conversations which should provide enough laughs for the whole weekend.

Quote:

---

Originally Posted by Overworked_PhD

Code:

---

{
 int x = 25;
}

---

is the same as

Code:

---

{
 int x; /*x initially has a garbage value*/
  x = 25;
}

---

---

Even if someone would grant that the above comparison is true (though I would not be inclined to do so, given the source), I find it funny how you are trying to BullSchildt your way out of this. What you are saying in the quote above, has nothing at all to do with that post of yours which started all of this (except for the x = 25 bit). What I mean is that even a child would understand the following:

Code:

---

#include <stdio.h>

int x = 3;

int main(void)
{
  printf("The value of x is: %d\n", x);

  {
    int x = 25;
    printf("The value of x is now : %d\n", x);
  }

  printf("The value of x again is: %d\n", x);

  return 0;
}

---

As 'x' is 3. Now 'x' changes to 25. Now 'x' is 3 again. If you had intended to say:

Quote:

---

'x' is 3. Now 'x' is some junk value which is changed to 25. 'x' is 3 again

---

why didn't you articulate it that way?

Quote:

---

Originally Posted by kermit

As 'x' is 3. Now 'x' changes to 25. Now 'x' is 3 again. If you had intended to say:

why didn't you articulate it that way?

---

It wouldn't matter if he had said it that way. He would still have been wrong in his output. If he wanted to be "technical", he would have said:

Code:

---

#include <stdio.h>

int x = 3;

int main(void)
{
  printf("The value of x is: %d\n", x);

  {
    int x = 25;
    printf("A new object named x has the value of: %d\n", x);
  }

  printf("The original object x still has the value of: %d\n", x);

  return 0;
}

---

But he didn't, because he didn't have even the remotest inkling of how object scope works.


Quzah.

Quote:

---

Originally Posted by kermit

Even if someone would grant that the above comparison is true (though I would not be inclined to do so, given the source), I find it funny how you are trying to BullSchildt your way out of this.

---

I've talked to Schildt. Even though his books suck, I can honestly say that he is light years ahead of your in terms of anything except book writing. The man in a brilliant Computer Scientist and 100x more articulate than what you or Quzah will ever be.

Quote:

---

Originally Posted by quzah

It wouldn't matter if he had said it that way. He would still have been wrong in his output. If he wanted to be "technical", he would have said:

Code:

---

#include <stdio.h>

int x = 3;

int main(void)
{
  printf("The value of x is: %d\n", x);

  {
    int x = 25;
    printf("A new object named x has the value of: %d\n", x);
  }

  printf("The original object x still has the value of: %d\n", x);

  return 0;
}

---

But he didn't, because he didn't have even the remotest inkling of how object scope works.


Quzah.

---

Uh... I don't think there is 'object scope'. I believe ANSI/ISO C would call this 'block scope.' But whatever. I'm getting used to your poor technical responses. With that. Someone with zero clue about how scope works wouldn't have come up with such an example.

Quote:

---

Originally Posted by Overworked_PhD

I've talked to Schildt. Even though his books suck, I can honestly say that he is light years ahead of your in terms of anything except book writing. The man in a brilliant Computer Scientist and 100x more articulate than what you or Quzah will ever be.

---

You know what man? I read one of your posts on the google groups and now I actually feel bad for you, and I really mean that, I am not being ironic. Yes, you know which one, the one about writing code with hate because of .... things that I will not share here.

The fact that you want to learn is admirable but the fact that you have no humility during this process and act presumptuous is not. When you say something that is false, people will contradict you, it's nothing personal. Take it as it is.

Quote:

---

Originally Posted by claudiu

You know what man? I read one of your posts on the google groups and now I actually feel bad for you, and I really mean that, I am not being ironic. Yes, you know which one, the one about writing code with hate because of .... things that I will not share here.

The fact that you want to learn is admirable but the fact that you have no humility during this process and act presumptuous is not. When you say something that is false, people will contradict you, it's nothing personal. Take it as it is.

---

Uh..... never mind. Tell me again why are just stuck to maintaining code at some hick firm no one has heard about.

Quote:

---

Originally Posted by claudiu

You know what man? I read one of your posts on the google groups and now I actually feel bad for you, and I really mean that, I am not being ironic. Yes, you know which one, the one about writing code with hate because of .... things that I will not share here.

The fact that you want to learn is admirable but the fact that you have no humility during this process and act presumptuous is not. When you say something that is false, people will contradict you, it's nothing personal. Take it as it is.

---

Also, I run under a different name on comp.lang.c you dolt.

Quote:

---

Originally Posted by Overworked_PhD

Uh..... never mind. Tell me again why are just stuck to maintaining code at some hick firm no one has heard about.

---

Ok, I see you don't want to take any good advice.