replace the "*" in the above string with " ", a single space character using ANSI C??

ok, I am very new to C programming and i have a problem at hand, i wuld really really appreciate any help plz,



The string is defined as the following:

char * foo = " Baltimore *MD*is*a*really*good*place*to*work*and*play";



Using the smallest amount of code possible, replace the "*" in the above string with " ", a single space character using ANSI C language. The resulting string should be printed out at the end of the program. In addition to a manual examination, the resultant .C file will be compiled with GCC compiler with the "-ansi" flag set and executed to determine success




CAN SOMEBODY HELP PLEASE?

Since this appears to be homework, I won't give you a solution.

What have you learned so far on how to go about it? Don't ask for a solution. Try it yourself and tell us where you're stuck.

On a side note, I think the people that wrote the assignment need to get a clue about string literals and read-only memory. :mad:

schools r i guess not started yet summer is still going on, and i do apologize that i got frustrated about it and posted in all C forums, i need it for my work, i would apreciate if someone gets back with some help, oh also i can see in the problem they r using pointers and i am soooooo not good at it, so any help?

It's an assignment of some sort. I doubt you have a job where the manager gives you a task like that, in such a pathetic manner. If it's work, consider quitting. ;)

So as I said before, what have you learned about C? Have you covered arrays, strings and such before?

and i did learn arrays and strings, but nothing on this advance level, i would love if you can point me somewhere where i can learn it like very quick? i know you are not giving out just whole solution, but you can atleast tell me where do i go and learn it

Duplicate threads deleted.

Quote:

---

Originally Posted by hardik

and i did learn arrays and strings, but nothing on this advance level, i would love if you can point me somewhere where i can learn it like very quick? i know you are not giving out just whole solution, but you can atleast tell me where do i go and learn it

---

If you know arrays, then why not walk through the string in a loop and change each char?

Disclaimer: I know the string in this example should be considered read-only, so a copy should be done and all alterations on the string should be done to the copy, but if the teacher/professor/manager/whatever wants it done this way, then whatever. Stupid, but whatever.

well, its not a teacher/professor/manager. lolz i see u still firmly believe its for school assignment, i thought of it very first but then i saw its asking smallest amount of code possible, dont you think it will be too long?

http://www.cprogramming.com/tutorial.html

Well, then if it's not for school, what is it?

Let me explain real quick what is wrong with what you got since it's still bothering me:

Code:

---

char *szHello1 = "Hello\n";
char szHello2[] = "Hello\n";

---

The first one is generally implemented into read-only memory on most systems even though I believe C makes no assumptions and allows you the opportunity to mess with the string. If you try to write to it, you may crash, you may not. What happens can't be guarenteed.

The second case allows you to manipulate the data because it implicitly makes a copy. That means it copies the string from read-only memory to the array that is properly sized to receive the string.

This is somewhat advanced, but you're asking how to overwrite the first string, which shouldn't be done. You need to either change the declaration to the second form or copy it over to a new buffer.

So anyway, back to your problem, no I don't think it will be too long to do that. What's your alternative? You want to use some other function that looks through the string just like your program will do? Only thing I could suggest would be if you already know the spacing of the '*''s, then you could handle it faster than a straight linear search.... ie, if every other char was a '*' then you would search every other char and replace.

For that specific string, if you want to be super fast, just change each position manually by index. You already know each index where a '*' is, so it's a simple assignment and can be done on one line.... ie. x = y = z = ' '; except you'd be doing it for each char of the string that has a '*'.

How's that for a hint?

lying is also bad, just because it's summer there doesn't mean it's summer here. If you've covered arrays, strings and conditional loops then there is no reason why you wouldn't be able to do this simple not advanced task.

strchr, loops and pointer dereferencing are your friends

Why bother with strchr? You can do it in a simple loop with a condition...

indeed cleaner to do it like that

I won't give you a solution. but i wanna help u.
MacGyver is Rite that "Do Your Own HomeWork".

if you fellow these step, then you will write your program by own.

Code:

---

#include <iostream.h>
#include <conio.h>
#include <string.h>

char * foo = " Baltimore *MD*is*a*really*good*place*to*work*and*play";

void main(){

        1) write a loop,
        2) check where '*' is present in String
        3) then replaced by ' '.

getch();
}

---

iostream.h is a deprecated C++ header file. conio.h is not a very portable library. If you can avoid it, and the getch() function, I would. Oh, and main() returns an int. ;)

Ya i known it. :)
but i worte my program for bigginer, not for Pro. :)

It's not valid C nor is it valid C++, beginner or not.

You also won't be able to legally nor safely edit foo (it's a string literal)

Quote:

---

char * foo = " Baltimore *MD*is*a*really*good*place*to*work*and*play";

---

foo is read-only as far as you're concerned. See MacGyver's post http://cboard.cprogramming.com/showp...2&postcount=14

Code:

---

#include <stdio.h>

int main(void)
{
    char s[] = "Baltimore *MD*is*a*really*good*place*to*work*and*play";
    /* Perhaps a loop could go here? That goes from 0 to strlen(s), or 0 to '\0' ? */
    return 0;
}

---

That's as about as much help as you're getting for the solution, if you have specific problems - ask.

i was trying to post my code so i can get further help, i couldnt figure how to do that, help plz?

I imagine you might be having trouble using code tags.....

This:

[code]x = 1337;[/code]

Produces:

Code:

---

x = 1337

---

insert

Code:

---

#include <string.h>

int main (void)
{
    char *foo = " Charlotte *NC*is*a*really*cool*place*to*work*and*play";
    char delims[]="*";
    char *result = NULL;
    result = strtok ( str, delims );
    while ( result != NULL )
    {
          printf ("result is "\&#37;s"\n", result );
          result = strtok (NULL, delims );
          }
          return 0;
          }

---

ok so now do i get some help, i wrote code, can someone help me correct it?
yeah macgyver thats exactly what was going wrong when i was trying to post my code, thanks

* Read all the posts in this thread, in order. My last post explains why you can't do that (with foo) - it's a string literal, and strtok modifies the string (which is illegal in this case)

* strtok doesn't help.

* Include stdio.h

What don't you understand? You seem to be ignoring what people have been saying. Start at the begging of foo, go through it in a loop, if the current character is '*' then set it to ' ' until you get to the end of foo (ie '\0'). The lastly print the result.

sorry guys, i am complete newbie so taking very long to understand what exactly you guys saying, well i didnt mean to ignore, trust me i read each reply throughout and tried to understand it, i really appreciate it. right now i am working on writing a code the way you said i should, i will be back with that new code

Okay this is as close as you'll get...

Code:

---

#include <stdio.h>
#include <string.h>

int main(void)
{
    char foo[] = " Charlotte *NC*is*a*really*cool*place*to*work*and*play";
    size_t  i = 0,
            len = strlen(foo);
    for(i = 0; i < len; i++)
    {
        /* foo[i] contains the current character, what should I do with it!?!? */
    }

    /* I should probably print foo here... */
    return 0;
}

---

Now it's up to you to complete it and trim it down, it can be written in less code (ie with less variables and less statements... I've merely shown you how to step through foo char-by-char nothing more, nothing less.

Code:

---

#include <stdio.h>
#include <string.h>

int main (int argc, char* argv[])
{
    char *foo = "Charlotte *NC*is*a*really*cool*place*to*work*and*play";
    char *bar = strdup (foo);
    char baz [] = "Charlotte *NC*is*a*really*cool*place*to*work*and*play";
    char *replace = baz;
    printf ("&#37;s\n", baz);
    while (*replace)
    {
        if (*replace == '*') *replace = ' ';
        replace++;
    }

    printf ("%s\n", baz);
        return 0;
}

---

hey zacs7, is this what you meant me to do? please tell me if there is anything wrong wiht it
or do you think the way you have done it with strlen is better? as in making code short or long?

It's almost correct.

* Remove foo & bar
* Try to be explicit with the loop, makes for a better read, ie.

Code:

---

while(*replace != '\0')
{
    if(*replace == '*')
        *replace = ' ';
/* and so on */

---

If you want to keep the original as well as the modified, just make a buffer the same size as baz and write the contents to that (in the loop, char-by-char replacing * with space as you go) just don't forget the null terminator. Otherwise you're done, welldone :D

The easiest solution ever.

Code:

---

#include <stdio.h>
char text[]={"Charlotte *NC*is*a*really*cool*place*to*work*and*play"};

int main (int argc, char* argv[]){
    int i;
    printf ("&#37;s\n", text);
    for(i=0;text[i];i++){
        if(text[i]=='*'){
            text[i]=' ';
        }
    }
    printf ("%s\n", text);
    return 0;
}

---

i have a question for maxotator and zacs7, maxotator, in your code wont we need secondary buffer? same for zacs7 if i remove foo and bar wont i need a secondary buffer? i tried and got an error, also in my code i used strdup because strcpy strcpy will not allocate the memory for the copy for you, can i replace it with something more simple? how is the memory allocated in your code maxorator? oh btw i am asking to understand it not to critisize it, thank you both ofyou.

here is rite code .

Code:

---

#include <stdio.h>
#include <string.h>

char * foo = " Baltimore *MD*is*a*really*good*place*to*work*and*play";

int main(void){

for (int i=0 ; i<strlen(foo) ; i++)
      if (foo[i]=='*')
          foo[i]=' ';

getch();
}

---

sorry this one is rite.

Code:

---

#include <stdio.h>
#include <string.h>

char * foo = " Baltimore *MD*is*a*really*good*place*to*work*and*play";

int main(void){

for (int i=0 ; i<strlen(foo) ; i++)
    if (foo[i]=='*')
        foo[i]=' ';

return 0;
}

---

try running it.

Lol, what's wrong with mine? :D

Quote:

---

Originally Posted by maxorator

Lol, what's wrong with mine? :D

---

Nothing, it's better than obaid's in the sense that using strlen as part of a for-loop end-check is a bad thing - especially when the string is being modified inside the loop, as the compiler may not realize if that the length is constant, and thus call strlen EVERY single iteration of the loop.

--
Mats

obaid's code is modifying a string literal too which are constant.

Why are there no standard restrictions on modifying string literals?

Quote:

---

Originally Posted by Smiley10

obaid's code is modifying a string literal too which are constant.

Why are there no standard restrictions on modifying string literals?

---

In trivial cases, it would be possible for the compiler to detect it, but in anything but trivial cases (e.g. when a function takes a char * as a parameter, and that char * happens to be a pointer to a literal).

It's not just obaid's code that does so, by the way, several of the other examples do too. Maxorators doesn't - but it also doesn't match the original poster's question - and by the way, we haven't seen any post from the OP for a while.... Perhaps the lesson ended? :-)

--
Mats

Quote:

---

Originally Posted by hardik

i have a question for maxotator and zacs7, maxotator, in your code wont we need secondary buffer? same for zacs7 if i remove foo and bar wont i need a secondary buffer? i tried and got an error, also in my code i used strdup because strcpy strcpy will not allocate the memory for the copy for you, can i replace it with something more simple? how is the memory allocated in your code maxorator? oh btw i am asking to understand it not to critisize it, thank you both ofyou.

---

The string is stored on the stack (as an array of characters), we step through the original string, changing each * for a space, therefore modifying the original string - meaning no 2nd buffer is required, but if you wanted to... you can store the modified string in another buffer don't copy it at the start, do it as you check for '*' (since it's redundant - you'll be looping twice without knowing it).

Quote:

---

Originally Posted by matsp

In trivial cases, it would be possible for the compiler to detect it, but in anything but trivial cases (e.g. when a function takes a char * as a parameter, and that char * happens to be a pointer to a literal).

It's not just obaid's code that does so, by the way, several of the other examples do too. Maxorators doesn't - but it also doesn't match the original poster's question - and by the way, we haven't seen any post from the OP for a while.... Perhaps the lesson ended? :-)

--
Mats

---

Thank you.

OP is back hehe, well my problem was solved, so BIG THANKS TO ALL YOU GUYS OUT THERE HELPING OTHERS SO MUCH, but also just for sake of learning i am still keeping up with all your posting its interesting how quick one can learn frm a forum like this rather than going thru some biggg lessons and still geting clueless. btw i found maxorator's code most simple and short, but hey smiley i would love if you can explain more throughout that in which case it would not work?

Code:

---


#include <stdio.h>
#include <string.h>

int main (int argc, char* argv[])
{
   
    char *foo = "Charlotte *NC*is*a*really*cool*place*to*work*and*play";
    char *bar = strdup (foo);
    char baz [] = "Charlotte
*NC*is*a*really*cool*place*to*work*and*play";
    char *replace = baz;
    printf ("&#37;s\n", baz);
    while (*replace)
    {
        if (*replace == '*') *replace = ' ';
        replace++;
    }

    printf ("%s\n", baz);
        return 0;
}

---

well maxotator, your code gives correct result but it changes the requirement, i think i would have to use the secondary buffer no matter what, as zacs said i am trying to remove foo and bar, and use char buf[100] as a secondary buffer, i will post back if i succeed, if anyone else can plz let me know

Okay this is as close as you'll get...

Code:

---

#include <stdio.h>
#include <string.h>

int main(void)
{
    char foo[] = " Charlotte *NC*is*a*really*cool*place*to*work*and*play";
    size_t  i = 0,
            len = strlen(foo);
    for(i = 0; i < len; i++)
    {
        /* foo[i] contains the current character, what should I do with it!?!? */
    }

    /* I should probably print foo here... */
    return 0;
}

---

Now it's up to you to complete it and trim it down, it can be written in less code (ie with less variables and less statements... I've merely shown you how to step through foo char-by-char nothing more, nothing less.




well wont i have to store foo in secondary buffer first? since i cant move/edit original one? what else can i use besides strdup to ease the process?

You can edit it if it's an array and not a pointer to the string literal. Because it's an array in this last example, you're fine in altering it. strdup()...... mmmm..... no.

Quote:

---

Originally Posted by obaid

sorry this one is rite.

Code:

---

#include <stdio.h>
#include <string.h>

char * foo = " Baltimore *MD*is*a*really*good*place*to*work*and*play";

int main(void){

for (int i=0 ; i<strlen(foo) ; i++)
    if (foo[i]=='*')
        foo[i]=' ';

return 0;
}

---

---

i tried running it, unsuccessful attempt tho

Quote:

---

i tried running it, unsuccessful attempt tho

---

Just ignore obaid for now. As everyone has been saying, he's trying to modify a string-literal, which is naughty and illegal, he also recalculates the string's size on every iteration - which is slow, as matsp said.

Code:

---

#include<stdio.h>
#include<string.h>
int main()
{
 char *foo = "Charlotte*NC*is*a*really*cool*place*to*work*and*play";
 char buf[100];
 int l = strlen(foo);
int i;

 for(i=0;i<l;i++)
 {
 
        if(*foo=='*')
  buf[i] = ' ';
  else
  buf[i] = *foo;
        *foo++;
 
 }
 for(i=0;i<l;i++)
  printf("%c",buf[i]);
  printf("\n");
  getchar ();
}

---

check it out guys, how u like it?

How about a 1 Liner:

Code:

---

#include<stdio.h>

int
main() {
 char fooBar[] = "Charlotte*NC*is*a*really*cool*place*to*work*and*play";
  char *p;

  for(p=fooBar; *p ; p++ )  if (*p=='*') *p = ' ';
 
  printf("%s\n",fooBar);

}

---

Or

Code:

---

    p = fooBar;
    while(p=strchr(p,'*') ) *p++='*';

---

Quote:

---

Originally Posted by hardik

Code:

---

#include<stdio.h>
#include<string.h>
int main()
{
 char *foo = "Charlotte*NC*is*a*really*cool*place*to*work*and*play";
 char buf[100];
 int l = strlen(foo);
int i;

 for(i=0;i<l;i++)
 {
 
        if(*foo=='*')
  buf[i] = ' ';
  else
  buf[i] = *foo;
        *foo++;
 
 }
 for(i=0;i<l;i++)
  printf("%c",buf[i]);
  printf("\n");
  getchar ();
}

---

check it out guys, how u like it?

---

Don't increase the foo pointer, or else you'll lose what it's pointing to. And if you null-terminate buf, you'll be able to print it as a string, ie

Code:

---

printf(buf)

---