I'm new taking my 1st class, and I am having trobble with my homework.. help please!

/*Homework2
Wednesday, September 21st, 2011 1:02 PM
Homework 2: The program asks user for a positve integer greater than 0 and less than 10. Program prints out:

1. whether number is even or odd
2. cube of number
3. square root of number
4. Sum of the digits
5. factorial

If an unsuitable number is entered, program exits only printing that an invalid number was entered.*/

Code:

---

#include <stdio.h>
#include <math.h> /* for sqrt*/


/* ask user to input a number between 1 and 10.
program will check number and terminate if input invalid
Otherwise the program will prnt out if the number is odd or even
value cubed, sqare root of the number, sum of the digits and factoral*/

int main()
{
    int number, i, sumx, productx, cubed, sqrt, factoral;
    printf(" enter a postive number between 1 and 10:");
    scanf("%d", &number);
    /* if ((number > 0 ) && (number < 11)).....*/
   
    { number/2 = even
    printf ("the number %d is even\n");
    else
    printf ("the number %d is odd\n"); /* 1 */
    }
   
   
    {cubed = (number*number*number)
    printf (" the number cubed is %d\n"); /*2*/
    }
   
    {sqrt = (sqrt(number))
    printf (" the number square root is %d\n"); /*3*/
    }
   
    {sumx=0;
    for (i=1;i<=number;i=i+1)

    sumx =sumx+ i;
    printf("the sum of the digits is %d\n", sumx); /* 4 */
   
    for (i=1; i<= number; i=i+1)
    productx=productx*i;
    }
   
    {factoral=0;
    for (i=1;i<=number;i=i*1)

    factoral =factoral* i;
    printf("factoral of this is %d\n", factoral); /* 5 */
   
    for (i=1; i<= number; i=i*1)
    factoral=factoral*i;
    }

  system("PAUSE");   
  return 0;
}

---

What is your question? Does the program work as expected or do you have some problem with it, if so what kind of problem?

it doesn't work... I can get only the sum of the numbers to work if I take all the rest out. I need the program to print 5 answers for each part of the assignment. the user enters if the user enters a number 1 though 10 then the program should print out answers to each of these questions...
1. whether number is even or odd
2. cube of number
3. square root of number
4. Sum of the digits
5. factorial

So lets say the user types in 2 I need it to output
2 is odd, the cube is 8 , square root is 1, sum of the digits is 3 and the factorial is 2

also dunno if this is a factor but I"m using dev C

Start by commenting out everything but the first part (odd/even), then add them in one after one until they are all working. Looking at the first part:

Code:

---

    /* if ((number > 0 ) && (number < 11)).....*/
   
    { number/2 = even
    printf ("the number %d is even\n");
    else
    printf ("the number %d is odd\n"); /* 1 */
    }

---

That is not even going to compile. The easiest way to determine if a number is uneven is to check if it's rightmost bit is set IMO.

what does it mean "it's rightmost bit is set IMO"? I don't think we learned that........

If you have not used bitwise operators then you haven't. But actually you could leave that bit since the first requirement (checking the range of the number) is commented out. So, solve that first. I would look for numbers < 0 and > 10 instead, if 'number' pass that test, print an error message and return 0; which will end the program.

I think I get what you mean... but I'm not sure how to do it correctly.... this is my attempt.....

Code:

---


int main()
{
    int number, i, sumx, productx, cubed, sqrt, factoral;
    printf(" enter a postive number between 1 and 10:");
    scanf("%d", &number);
    if (number > 0);
    printf (" number is not between 1 and 10\n");
    else
    if (number < 11);
    printf (" number is not between 1 and 10\n");
   
    /* if ((number > 0 ) && (number < 11)).....*/

---

thanks
-Cess

Quote:

---

Originally Posted by Cess

what does it mean "it's rightmost bit is set IMO"? I don't think we learned that........

---

In an integer value the least significant (right most) bit is always 1 for odd numbers and 0 for even numbers... it's a side effect of computers working in binary.

Code:

---

if ((number & 1) > 0)
  printf("%d is odd",number);
else
  printf(%d is even", number);

---

You can also use

Code:

---

if ((number % 2) > 0)
  printf("%d is odd",number);
else
  printf(%d is even", number);

---

Subsonics gave you excellent advice about working through it one small bit at a time. That's how we "real" programmers do things... write a given function or section of code and test... work on the next, test again... until it's all working.

The computer does things in little tiny steps, one at a time... and so should you.

Quote:

---

Originally Posted by Cess

I think I get what you mean... but I'm not sure how to do it correctly.... this is my attempt.....

Code:

---


int main()
{
    int number, i, sumx, productx, cubed, sqrt, factoral;
    printf(" enter a postive number between 1 and 10:");
    scanf("%d", &number);
    if (number > 0);
    printf (" number is not between 1 and 10\n");
    else
    if (number < 11);
    printf (" number is not between 1 and 10\n");
   
    /* if ((number > 0 ) && (number < 11)).....*/

---

thanks
-Cess

---

Code:

---

if ((number < 1) || (number > 10))
 {
    printf("I specifically said from 1 to 10\n");
    exit(1);
  }

  // rest of program here

---

If the number is less than 1 OR the number is greater than 10 the program exits.

> printf ("the number %d is even\n");
Same mistake as in another thread - you're not supplying a parameter to match the %d

Mind you, you're using a gcc compiler, so perhaps you can find where in dev-c++ you set the compiler flags.

Add -W -Wall -ansi -pedantic and then see more error messages to fix.

Quote:

---

Originally Posted by Cess

I think I get what you mean... but I'm not sure how to do it correctly.... this is my attempt.....

Code:

---


int main()
{
    int number, i, sumx, productx, cubed, sqrt, factoral;
    printf(" enter a postive number between 1 and 10:");
    scanf("%d", &number);
    if (number > 0);
    printf (" number is not between 1 and 10\n");
    else
    if (number < 11);
    printf (" number is not between 1 and 10\n");
   
    /* if ((number > 0 ) && (number < 11)).....*/

---

thanks
-Cess

---

You have the answer right at the end there, but you need to change it a bit. 'number' can not be both less than 1 and larger than 10, so use OR instead of AND, that is || so:

Code:

---

if( (number < 1) || (number > 10) ) {
    printf("The number is out of range. (1-10)\n");
    return 0;
}

---

Ok tried fixing it and still getting error.... We haven't used || before so I don't really know how to use it but ya this is what I did and got an error on the return line it says....

Code:

---


#include <iostream>
using namespace std;
int main()
    {
    int number;
    printf(" enter a positive number between 1 and 10:");
    if( (number < 1) || (number > 10) ) {
    printf("The number is out of range. (1-10)\n");
   
system("PAUSE");
return 0; }

---

You're going to get an error alright... you're using C++ headers and C doesn't have the first clue what "namespace" is.

Plus your brackets are mismatched ...

You need to adopt a good indentation style. Watch:

Code:

---

#include <iostream> // this is C++
using namespace std; // so is this
int main() // this should really be (void) instead of just ()
{
    int number;
    printf(" enter a positive number between 1 and 10:");
    if( (number < 1) || (number > 10) ) {
        printf("The number is out of range. (1-10)\n");
   
        system("PAUSE");
        return 0; }
// where is the matching }

---

Count the pairs of braces.


Quzah.

Quote:

---

Originally Posted by Cess

also dunno if this is a factor but I"m using dev C

---

No you're not... you're using DevC++ ... a whole different language. There is no Dev C.

Also you should be aware that compiler and IDE have been abandonware for most of a decade. The compiler is hopelessly out of date and the guy who wrote the IDE is noplace to be found.

I think, you should give serious consideration to getting onto a purely C compiler, C-99 compliant, and a new IDE that is up to date...
I generally suggest Pelles C but there are others you can use.

Double Header! ... Hi Quzah...

ops... that was silly of me...that happen cuz I started over and it automatically had that part and I didn't double check it sorry..... I'm extremely new so ya sorry.
lets try this again..... why am I still getting errors...

Code:

---

#include <stdio.h>
#include <math.h>

int main()
{   
    {int number;
    printf(" enter a positive number between 1 and 10:");
    }
    {if( (number < 1) || (number > 10) )
    printf("The number is out of range. (1-10)\n");
    }
system("PAUSE");
return 0;
}

---

Quote:

---

Originally Posted by Cess

ops... that was silly of me...that happen cuz I started over and it automatically had that part and I didn't double check it sorry..... I'm extremely new so ya sorry.
lets try this again..... why am I still getting errors...

---

Because you aren't paying attention to what you are doing, and are really just throwing random bits of code in trying to get lucky:

Code:

---

#include <stdio.h>
#include <math.h>

int main()
{   
    {int number;
    printf(" enter a positive number between 1 and 10:");
    }
    {if( (number < 1) || (number > 10) )
    printf("The number is out of range. (1-10)\n");
    }
system("PAUSE");
return 0;
}

---

More does not mean better. Everything inside a pair of braces goes out of scope as soon as you hit the closing brace. All the inner vanishes when you hit the closing braces.


Quzah.

Quote:

---

Originally Posted by Cess

ops... that was silly of me...that happen cuz I started over and it automatically had that part and I didn't double check it sorry..... I'm extremely new so ya sorry.
lets try this again..... why am I still getting errors...

Code:

---

#include <stdio.h>
#include <math.h>

int main()
{   
    {int number;
    printf(" enter a positive number between 1 and 10:");
    }   
    {if( (number < 1) || (number > 10) )
    printf("The number is out of range. (1-10)\n");
    }system("PAUSE");
return 0;
}

---

---

Because you are using the rules of scope incorrectly...

Code:

---

    {int number;    <--- you start a new scope here
    printf(" enter a positive number between 1 and 10:");
    }                      <-- you end it here, number is destroyed.
    {if( (number < 1) || (number > 10) )    <--- you start another scope here but number is unknown
    printf("The number is out of range. (1-10)\n");
    }                                        <--- you end the second scope here

---

Yes, that's right, those curly braces are active parts of the software, not merely window dressing.

try this...

Code:

---

#include <stdio.h>
#include <math.h>

int main()
{   
    int number;
    printf(" enter a positive number between 1 and 10:");

scanf("%d",&number);
   
    if( (number < 1) || (number > 10) )
        printf("The number is out of range. (1-10)\n");
   

return 0;
}

---

AND another double header... Hi again Quzah!

Quote:

---

Originally Posted by CommonTater

AND another double header... Hi again Quzah!

---

:eyebrow: No wonder your post count is so high!


Quzah.

Quote:

---

Originally Posted by quzah

:eyebrow: No wonder your post count is so high!
Quzah.

---

LOL... Point taken.

ya I know I'm probably annoying I'm I am trying to learn and I really don't know what I am doing ok tried fixing it again...and I failed
something wrong with the pause... dunno

Code:

---

#include <stdio.h>
#include <math.h>

int main()
{   
    int number;
    printf(" enter a positive number between 1 and 10:");

scanf("%d",&number);
   
    if( (number < 1) || (number > 10) )
        printf("The number is out of range. (1-10)\n");
   
system("PAUSE");
return 0;
}

---

Quote:

---

Originally Posted by Cess

ya I know I'm probably annoying I'm I am trying to learn and I really don't know what I am doing ok tried fixing it again...and I failed
something wrong with the pause... dunno

Code:

---

#include <stdio.h>
#include <math.h>

int main()
{   
    int number;
    printf(" enter a positive number between 1 and 10:");

scanf("%d",&number);
   
    if( (number < 1) || (number > 10) )
        printf("The number is out of range. (1-10)\n");
   
system("PAUSE");
return 0;
}

---

---

When stuff like that happens... look up the keyword in your C library documentation... (which is exactly what I just did)

In this case the problem is that you need to #include <stdlib.h> or it doesn't know what system() is.

Quote:

---

Originally Posted by CommonTater

When stuff like that happens... look up the keyword in your C library documentation... (which is exactly what I just did)

In this case the problem is that you need to #include <stdlib.h> or it doesn't know what system() is.

---

I did this exact thing when I wrote that calculator earlier. ;)

I know where a bunch of functions are from, but I don't have them all memorized. What's the point, when I can make a new tab, go up to my URL bar and type "man system(3)" and have it pull up the man page for me?


Quzah.

Quote:

---

Originally Posted by quzah

I did this exact thing when I wrote that calculator earlier. ;)

I know where a bunch of functions are from, but I don't have them all memorized. What's the point, when I can make a new tab, go up to my URL bar and type "man system(3)" and have it pull up the man page for me?
Quzah.

---

With over a thousand functions in the library, nobody is going to memorize even half of them... and it only gets worse in Windows API where there are over 30,000 functions... Knowing how to look stuff up is probably the programmer's second most important skill.

And ... :tongue: ... in Pelles C all you have to do is press F1.

ok I'm getting closer... I got 1 now I'm on number 2 and I need help again Lol thank you guys for being patient and helpful to a newbie like me!

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
 

int main()
{   
    int number;
    printf(" enter a positive number between 1 and 10:");
    scanf("%d",&number);
   
    if( (number < 1) || (number > 10) )
        printf("The number is out of range. (1-10)\n");
   
    /* Check to see if number is odd or even #1 */
    if ((number & 1) > 0)
      printf("%d is odd\n",number);
      else
      printf("%d is even\n", number);
   
      /* cube the number #2 */
      number*number*number
      printf("%d is the cube of that number\n");
     
   
system("PAUSE");
return 0;
}

---

You need to assign the value with =, which you aren't doing, and you don't even have it compilable at the moment:

Code:

---

//cube number
number = number * number * number;

---

Quzah.

Ok... so the number is out of range... what do you do about that?
hint: you don't let it go crashing down into the rest of your program, causing even more problems.

number*number*number does exactly nothing... the compiler should report it as a syntax error.

Your last printf might read better as

Code:

---

printf("%d^3 = %d\n", number, number * number * number);

---

Quote:

---

Originally Posted by quzah

You need to assign the value with =, which you aren't doing, and you don't even have it compilable at the moment:

Code:

---

//cube number
number = number * number * number;

---

Quzah.

---

Which is going to alter the value of number for the next function.

Quote:

---

Originally Posted by CommonTater

Which is going to alter the value of number for the next function.

---

That's true. I was only solving the immediate problem at hand. Your in-printf solution is more elegant.


Quazh.

Quote:

---

Originally Posted by Cess

ok I'm getting closer... I got 1 now I'm on number 2 and I need help again Lol thank you guys for being patient and helpful to a newbie like me!

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
 

int main()
{   
    int number;
    printf(" enter a positive number between 1 and 10:");
    scanf("%d",&number);
   
    if( (number < 1) || (number > 10) )
        printf("The number is out of range. (1-10)\n");
   
    /* Check to see if number is odd or even #1 */
    if ((number & 1) > 0)
      printf("%d is odd\n",number);
      else
      printf("%d is even\n", number);
   
      /* cube the number #2 */
      number*number*number
      printf("%d is the cube of that number\n");
     
   
system("PAUSE");
return 0;
}

---

---

Wasn't you also suppose to quit, if a number isn't 1-10? Otherwise you are just carrying on as if nothing happened.

Quote:

---

Originally Posted by Cess

If an unsuitable number is entered, program exits only printing that an invalid number was entered.*/

---

Code:

---

    if( (number < 1) || (number > 10) ) {
        printf("The number is out of range. (1-10)\n");
        return 0;
    }

---

Quote:

---

Originally Posted by quzah

That's true. I was only solving the immediate problem at hand. Your in-printf solution is more elegant.


Quazh.

---

Be still my fluttering heart... A compliment from Quzah ... now when's the last time that happened :D

you are totally correct I didn't realize that thanks... I'm working on square root right now but I will go back and try and fix that.

this is what I have for square root

Code:

---

/* square root the number #3 */
      sqrt=sqrt(number);
      printf("%d is the square root of that number\n",sqrt);

---

Quote:

---

Originally Posted by Cess

you are totally correct I didn't realize that thanks... I'm working on square root right now but I will go back and try and fix that.

this is what I have for square root

Code:

---

/* square root the number #3 */
      sqrt=sqrt(number);
      printf("%d is the square root of that number\n",sqrt);

---

---

Nope... look how I did it in message #30 ... copy the form of that, using the new math.

Also note that sqrt() is for floating point numbers and you're working with integers.
(You really gotta get into the habit of looking this stuff up!)

And, you can't use function names as variables...

there is something wrong again... and I can't figure out why its doing it....

this is my program so far....

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
 

int main()
{   
    int number, cube, sqrot ,i ,sumx ,fact ;
    printf(" enter a positive number between 1 and 10:");
    scanf("%d",&number);
   
    if( (number < 1) || (number > 10) ) {
        printf("The number is out of range. (1-10)\n");
        return 0;
        }
   
    /* Check to see if number is odd or even #1 */
    if ((number & 1) > 0)
      printf("%d is odd\n",number);
      else
      printf("%d is even\n", number);
   
      /* cube the number #2 */
      cube=number*number*number;
      printf("%d is the cube of that number\n",cube);
     
      /* square root the number #3 */
      sqrot=sqrt(number);
      printf("square root of that is %d\n",sqrot);
     
      /* Sum of the digits #4 */
      sumx=0;
    for (i=1;i<=number;i=i+1)
    sumx =sumx + i;
    printf("the sum of the digits is %d\n", sumx);

    /* Factorial of the number #5 */
    fact=0;
    for (i=1;i<=number;i=i+1)
    fact=fact*i
    printf("the factorial of that number is %d\n",fact);
   
system("PAUSE");
return 0;
}

---

it works fine until I added the last part to get it to do the factoral

Code:

---

  /* Factorial of the number #5 */
    fact=0;
    for (i=1;i<=number;i=i+1)
    fact=fact*i
    printf("the factorial of that number is %d\n",fact);

---

when that is added it jumps back up to the sqrot and says it has an error? Grrr I will get this program right it might take me all night but ya!!!! thanks again for help.

Ok... what EXACTLY is it doing or not doing?
What error messages are you getting?
What warnings/errors are your compiler listing?

All this is good information... "It doesn't work" tells us nothing.

You are missing a semicolon after fact*i. :)

true it says
In function 'int main()';
line 28 waring converting int to double
expected' before "printf"
[build error] error 1

I'm guessing its because I didn't use a double of the sqrt which is weird because it worked before and only stopped working when I added the factorial. Can I use int and double for the number they input? If so how do I do that correctly so it works the way I want it to?
-thanks again Cess

lol I can't believe I missed that simple ;

shoot
its sayin the factorial is zero.... so I'm still doing something wrong
I need it to give the factorial of the person's input

Code:

---

/* Factorial of the number #5 */
    fact=0;
    for (j=number;j<=number;j=j+1);
    fact=fact*j;
    printf("%d is the factorial\n",fact);

---

You initialize fact to zero, then you multiply that with i. What happens when you multiply something with zero?

ok now I'm getting 6....I don't understand how I am spost to set it up differently

Code:

---

  /* Factorial of the number #5 */
    fact=number;
    for (j=1;j<=number;j=j+1);
    fact=fact*j;
    printf("%d is the factorial\n",fact);

---

and I get 3 when I entered 2 and its spost to be 2....

Code:

---

/* Factorial of the number #5 */
    fact=1;
    for (j=1;j<=number;j=j+1);
    fact=fact*j;
    printf("%d is the factorial\n",fact);

---

It's the semicolon again, your for loop runs empty. what you have is the same as:

Code:

---

for(j = 1; j <= number; j=j+1)
    ;

---

I am still not understanding.....

Quote:

---

Originally Posted by Cess

I am still not understanding.....

---

A semicolon terminates a statement. Loops need a statement attached, typically, to actually do something:

Code:

---

#include<stdio.h>
int main( void )
{
    int x = 0;
    for( x = 0; x < 10; x++ )
        printf( "x is %d\n", x );
    return 0;
}

---

Run that. Now add a ; to the end of the line with the for loop. Run it again.

Without a pair of braces, only the immediate statement after a loop (or an if, or an else, etc.,) is attached:

Code:

---

if( x == 0 )
    ; // this statement is attached to the if
; // this one is not

---

Note, it doesn't matter what line you have, but what actual statement is next:

Code:

---

if( x == 0 )



    ; // this is attached to the if

---

Code:

---

if( x == 0 ); // this ; is attached to the if

---

Quzah.

Quote:

---

Originally Posted by Cess

I am still not understanding.....

---

A for loop have these components

Code:

---

for (i = 0; i < 10; i++) {
    /* loop body */
}

---

A semicolon in there will be interpreted as being part of the loop body, i.e an empty statement.

Quote:

---

Originally Posted by Subsonics

A for loop have these components

Code:

---

for (i = 0; i < 10; i++) {
    /* loop body */
}

---

A semicolon in there will be interpreted as being part of the loop body, i.e an empty statement.

---

To illustrate one step further:

Code:

---

for (i = 0; i < 10; i++) ; {
    /* not the loop body */
}

---

That ; makes all the difference.


Quzah.

Quote:

---

Originally Posted by Cess

ok I now understand that....but I don't think that was the only thing I had wrong....

I've tried chancing it a few times.... and I'm not getting it to give me the factorial...

Code:

---

    /* Factorial of the number #5 */
      fact=number;
      for (m=0;m<=number;m=m+1)
          {
          fact=fact*m;
          printf("%d is the factorial\n",fact);
          }

---

---

So now you are back to:

Quote:

---

Originally Posted by Subsonics

You initialize fact to zero, then you multiply that with i. What happens when you multiply something with zero?

---

It's not a race. Take your time and read what people are telling you.


Quzah.

ok I now understand that....but I don't think that was the only thing I had wrong....

I've tried chancing it a few times.... and I'm not getting it to give me the factorial...

Code:

---

  fact=number;
      for (m=1;m<=number;m=m+1)
          {
          fact=fact*m;
          printf("%d is the factorial\n",fact);
          }

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Attachment 10902

It's pretty obvious it's not the only thing you have wrong, when your output is "1 is the square root of...".


Quzah.

HA HA! Yeaah!! I got it to work.... thank you all for your help!!!!:biggrin:

Code:

---


int main()
{   
    int number, cube, i ,sumx ,fact ,m;
    float sqrot;
    printf(" enter a positive number between 1 and 10:");
    scanf("%d",&number);
   
    if( (number < 1) || (number > 10) ) {
        printf("The number is out of range. (1-10)\n");
        return 0;
        }
   
    /* Check to see if number is odd or even #1 */
    if ((number & 1) > 0)
      printf("%d is odd\n",number);
      else
      printf("%d is even\n", number);
   
      /* cube the number #2 */
      cube=number*number*number;
      printf("%d is the cube of that number\n",cube);
     
      /* square root the number #3 */
      sqrot=sqrt(number);
      printf("%12.5f is the square root of that number \n",sqrot);
     
      /* Sum of the digits #4 */
      sumx=0;
      for (i=1;i<=number;i=i+1)
      sumx =sumx + i;
      printf("%d is the sum of the digits \n", sumx);

      /* Factorial of the number #5 */
      fact=1;
      for (m=1;m<=number;m=m+1)
          {
          fact=fact*m;
          }
          printf("%d is the factorial\n",fact);
   
system("PAUSE");
return 0;
}

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Quote:

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Originally Posted by quzah

It's pretty obvious it's not the only thing you have wrong, when your output is "1 is the square root of...".


Quzah.

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You are being SO picky Quzah! :D :D

Quote:

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Originally Posted by Adak

You are being SO picky Quzah! :D :D

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He paid me a compliment and even admitted i had a better solution a bit earlier... I'm thinking he's just evening up the score :D