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what the... C code problem

Hi all,

I clearly am suffering from a fundemental lack of understanding. What is wrong with the following code snippet?

Code:

#define _GNU_SOURCE

#include <stdio.h>

#include <string.h>

#include <malloc.h>

#include <unistd.h>



int main( int argc, char **argv) {



        char *a;

        int b = 100;



        a = ( char * ) malloc( 4 );

        strcpy( a, ( char * ) b );

        printf( "%s\n", a );

        return ( 0 );

}

I get a segmentation fault. Do I not have to malloc memory for the point to char? Is four bytes not enough? Do I not have to cast b as a pointer to char in order to strcpy it into a? Or am I approaching storing/concatenating an int into a string in the wrong way?

TIA.

strcpy( a, ( char * ) b );

This isn't doing what you think it's doing. Its attempting to access memory location 100 and copy that to a. Which is why you have the segmentation fault.

sscanf( a, "%d", b );

Hi,

Thanks. You were right--that wasn't doing at all what I thought it was.
When I try sscanf, I get a warning:

junk.c:15: warning: format argument is not a pointer (arg 3)

And then the printf prints just the newline.

Ideas?

Thanks again.

I think you need this:

Code:

a=(char *)malloc(4);

sprintf(a,"%d",b);

...

Yep Matir is right mixing up my scan and my print.

Thank you both, all fixed now and ready to roll for a while!

Thanks again!

No problem. Feel free to click the thanks button by my post... not that I know what it does, lol. :)