Thread: Is there a standard function?

Hello everybody,

I recently started studying C, and so far I didn't find a function to format
integer numbers with thousand comma separators:

30000000 value to format
30,000,000 formatted value.

Do you know if there is a standard function to do this kind of formatting?

Thank you

There is no standard function to do that as far as I know but you can read here how to do it manually, although it may be a little too much to handle if you have just started learning the language.
Question 12.11

I tried to assemble a routine for the formatting task, but it looks like
I missed something:

Code:

// ----------------------------------------------------------------------------------------------
// Prog_name: comma_sep.c / ver 0.0
// Date: 27 june 2010
// Author: frktons @ cprogramming forum.
//-----------------------------------------------------------------------------------------------
// A routine for separating integer numbers with thousands comma separator. 
// Created with Pelles C for Windows 6.00.4
//-----------------------------------------------------------------------------------------------
#include <stdio.h>
#undef getchar
#pragma comment(lib, "\\masm32\\lib\\msvcrt.lib")

int main(int argc, char *argv[])
{
    int num1 = 30000;
    int num2 = 50000;
    int result = (num1 * num2);
    int num = result;
    char buffer[14] = {' '};
    int len_str = 14;
    int x;
    int remain = 0;
    int count = 0;
    char digit[10] = {'0','1','2','3','4','5','6','7','8','9'};    


    printf("\nThe value of result is: %d\n",result);

    for (x = len_str; x > 0; x--)
      {
	if (num == 0) 
          break;	

	if (count == 3)
	  {
	    count = 0;
	    buffer[x] = ',';
	    x--;
	  }

       remain = num % 10;
       num = num / 10;
       buffer[x] = digit[remain];
       count++;
    
      }  

    printf("\nThe formatted value of result is: %s\n",buffer);
    getchar();
    return 0;
}

What did I miss?

Thanks

Originally Posted by User Name:

I don't know if you just prefer //s or something, but you may find /**/s helpful.

What are you missing? What's wrong with the output?

Well, // and /* */ are quite the same in C99, no problem with that.

What's missing? the printf of the final string array doesn't show up.

Code:

    printf("\nThe formatted value of result is: %s\n",buffer);

Originally Posted by frktons

What did I miss?

Thanks

A C string is formally defined as a sequence of characters ending with and including zero. If you are going to fill a string backwards you have to take care to do it correctly:

Code:

int next , x;

for (x = len_str - 1 , next = 0; x > -1 && next < len_str; x-- ) {
   ... determine grouping ...

   if (comma) {
      buffer[x - next] = ',';
      next++;
   }

   buffer[x - next] = digits[remain];
   next++;

   ... update num ...
   
   if (num == 0) {
      buffer[next] = '\0';
      break;
   }
}

No nudging of x should be necessary; let the loop count backward.

While, in the program you wrote, result will never be negative, it is also impractical to think that always. You'll have to tweak it a bit for that.

Originally Posted by whiteflags

A C string is formally defined as a sequence of characters ending with and including zero. If you are going to fill a string backwards you have to take care to do it correctly:

Code:

int next , x;

for (x = len_str - 1 , next = 0; x > -1 && next < len_str; x-- ) {
   ... determine grouping ...

   if (comma) {
      buffer[x - next] = ',';
      next++;
   }

   buffer[x - next] = digits[remain];
   next++;

   ... update num ...
   
   if (num == 0) {
      buffer[next] = '\0';
      break;
   }
}

No nudging of x should be necessary; let the loop count backward.

While, in the program you wrote, result will never be negative, it is also impractical to think that always. You'll have to tweak it a bit for that.

Thanks whiteflags.

I have to meditate for a while in order to understand the syntax of your code,
in the meanwhile I came up with a crumpled working solution:

Code:

// ----------------------------------------------------------------------------------------------
// Prog_name: comma_sep.c / ver 0.0
// Date: 27 june 2010
// Author: frktons @ cprogramming forum.
//-----------------------------------------------------------------------------------------------
// A routine for separating integer numbers with thousands comma separator. 
// Created with Pelles C for Windows 6.00.4
//-----------------------------------------------------------------------------------------------
#include <stdio.h>
#undef getchar
#pragma comment(lib, "\\masm32\\lib\\msvcrt.lib")

int main(int argc, char *argv[])
{
    int num1 = 30000;
    int num2 = 50000;
    int result = (num1 * num2);
    int num = result;
    char buffer[14] = {' '};
    int len_str = 13;
    int x;
    int remain = 0;
    int count = 0;
    char digit[10] = {'0','1','2','3','4','5','6','7','8','9'};    


    printf("\nThe value of result is: %d\n",result);

    for (x = len_str; x > 0; x--)
      {
	if (num == 0) 
          break;	

	if (count == 3)
	  {
	    count = 0;
	    buffer[x] = ',';
	    x--;
	  }

       remain = num % 10;
       num = num / 10;
       buffer[x] = digit[remain];
       count++;
    
      }  

    printf("\nThe formatted value of result is: ");
    for (x = 0; x < 14; x++)
	printf("%c",buffer[x]);	 
    getchar();
    return 0;
}

And the output:

Code:

The value of result is: 1500000000

The formatted value of result is:  1,500,000,000

You are totally right about the number being always positive, and I'm
going to complete the logic of the code, as soon as I know how to do it.

This is only a simple exercise I'm doing in order to learn some C syntax, nothing
more for the time being. :-)

Actually, I'm lucky you're still here cause I made a mistake. Instead of using x, you should be using the len_str variable. We don't even need x:

Code:

int next;

for (next = 0; next < len_str; ) {
   ... determine grouping ...

   if (comma) {
      buffer[len_str - next - 1] = ',';
      next++;
   }

   buffer[len_str - next - 1] = digits[remain];
   next++;

   ... update num ...
   
   if (num == 0) {
      buffer[next] = '\0';
      break;
   }
}

Sorry for the inconvenience.

>>#undef getchar
getchar is a function, not a macro. Therefore, you cannot undefine it. The line is pointless.

>>#pragma comment(lib, "\\masm32\\lib\\msvcrt.lib")
This is implicit. The linker will always link to the standard C runtime library by default. It should be safe to remove this.

You point me to a source that says it's a macro?

But some compilers have macro and function versions
So undef line is not necessarily pointless.
Eg
IBM...
MSDN
Edit: sorry about double post. Internet's slow here... ;(

MSVC AFAIK does not have macro versions. But whatever floats your boat, I suppose.