Simple C Program that sometimes crashes!

**logitechz** · 12-03-2011

Hello everyone, I am having trouble understanding why "12345" is not being displayed and instead the program either displays nothing or it will crash. Helps!

Code:

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


int main()
{
    char* ch =0;  /* A nice charstar to hold 12345 as a string */
    int num=12345; /*12345 as a numerical value rather than a string */
    sprintf(ch,"%i",num); /*To convert the integer num into a string and store in it charstar! */
    printf("%s\n",ch); /*I want to see the results */
    return 0;
}

**grumpy** · 12-03-2011

You have initialized ch as the NULL pointer. Writing any data to it (which is what you are doing with sprintf()) gives undefined behaviour.

You need to make it point at a buffer long enough to handle the content you are writing to it.

Code:

    char ch[6];

will do the trick (assuming you don't try to write more than a 5-digit number to it: sprintf() adds a trailing zero terminator).

More generally: a pointer is not an array.

**logitechz** · 12-03-2011

Thanks for your help, so what you are saying is, I must define the size of the string I want to store in the char* variable before I actually try to write anything to it and never set char* to 0 aka NULL.

**TheBigH** · 12-03-2011

Yes, that's right.

**grumpy** · 12-03-2011

Originally Posted by TheBigH

Yes, that's right.

Only in the very loosest and most imprecise sense.

Have a look at this link for one guy's attempt to explain the differences between pointers and arrays.

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