Initializing a char array to be NULL strange behaviour

Question

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

int main(int argc, char *argv[])
{
        char compliance[256] = {'\0'};
        if(compliance == NULL)
        {
                printf("compliance is null \n");
                return 0;
        }
        printf("length of compliance %zd \n",strlen(compliance));
        return 0;
}

Output:

length of compliance 0


int main(int argc, char *argv[])
{
        char compliance[256] = {'\0'};
        memset(compliance,0,256);

        if(compliance == NULL)
        {
                printf("compliance is null \n");
                return 0;
        }
        printf("length of compliance %zd \n",strlen(compliance));
        return 0;
}

Output length of compliance 0

As many of you have pointed out I wanted to use memset (instead of memcpy).But still don't get why in the second program compliance is not NULL? or in other words How do I make it NULL?

I'm quite certain you want to use memset, not memcpy, just as i'm sure It is entirely unneeded given your initializer presence. — WhozCraig
– WhozCraig, Commented Nov 26, 2013 at 1:19

ensc · Accepted Answer · 2013-11-26 01:18:49Z

3

both programs are broken;

    if(compliance == NULL)

makes no sense as compliance is never NULL (it is a variable on stack)

In the second part

    memcpy(compliance,0,256);

copies from source address 0 (NULL) which causes segfault on most platform. You probably want to use memset here

answered Nov 26, 2013 at 1:18

ensc

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3 Comments

Glenn Teitelbaum Over a year ago

the first line will get optimized out and looks very much like a debugging statement to help remove that case from being the reason for the core dump - the second, ....

liv2hak Over a year ago

I didn't understand you first comment ? (compliance == NULL) why doesn't it make sense?

ensc Over a year ago

because compliance is an array and it is ensured that it is not at a NULL address. You might want char *compliance = malloc(256) or so.

Yu Hao · Accepted Answer · 2013-11-26 01:19:28Z

2

compliance is an array, not a pointer (An array name are automatically converted to a pointer to the first element in some situations, but arrays are not pointers), it will never be equal to a null pointer.

The segmentation fault in the second example is caused by the call to memcpy.

memcpy(compliance,0,256);

You are copying from a null pointer. Probably what you want is memset.

answered Nov 26, 2013 at 1:19

Yu Hao

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Comments

Glenn Teitelbaum · Accepted Answer · 2013-11-26 01:19:23Z

1

you probably meant memset

you are copying from address 0 - 256 bytes

answered Nov 26, 2013 at 1:19

Glenn Teitelbaum

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