Question

Please see the following code:

char h[256];
    h[0]=NULL;
    if(h!=NULL)
    {
        printf("It doesn't show NULL\n");
    }
    else
    {
        printf("It shows NULL\n");
    }

and also the following:

char h[256];
        if(h!=NULL)
        {
            printf("It doesn't show NULL\n");
        }
        else
        {
            printf("It shows NULL\n");
        }

and also the following:

 char h[256];
h[0]='\0';
            if(h!=NULL)
            {
                printf("It doesn't show NULL\n");
            }
            else
            {
                printf("It shows NULL\n");
            }

In every case the char* h doesn’t have NULL. Why is it the case? Isn’t it suppose to have NULL as I’m not storing anything there? And if it’s not the case, how I can I make sure that it contains nothing but NULL?

Answer 1

h[0] is not the same as h. h[0] is the first character in the array; h is the array itself (which decays to a pointer in this situation).

Try this:

char h[256];
h[0]=NULL;
if(h[0]!=NULL)
{
    printf("It doesn't show NULL\n");
}
else
{
    printf("It shows NULL\n");
}

Note also that you probably shouldn’t be using NULL in this situation. NULL is for pointers; you want '\0' instead.