why assert(str!=NULL) does not return an error?-CodePudding

Why did this not print an assert error?

This is my code:

#include<assert.h>
#include<stdio.h>

int main(){
    int n =12;
    char str[50] = "";
    assert(n>=10);
    printf("output :%d\n",n);
    assert(str!=NULL);
    printf("output :%s\n",str);
}

CodePudding user response：

char str[50] = "";

This makes str into an array of 50 chars, and initialises the memory to all zeroes (first byte explicitly from "" and rest implicitly, because C does not support partial initialisation of arrays or structs).

assert(str!=NULL);

When used in an expression, array is treated as pointer to its first element. The first element of the array very much has an address, so it is not NULL.

If you want to test if the first element of the array is 0, meaning empty string you need

assert(str[0] != '\0');

You could compare to 0, or just say assert(*str);, but comparing to character literal '\0' makes it explicit to the reader of the code, that you are probably testing for string-terminating zero byte, not some other kind of zero, even if for the C compiler they're all the same.

CodePudding user response：

str is an array which is effectively a pointer, which means the address in memory where the array is stored. A pointer is NULL if it doesn't point to any specific place in memory. Your str pointer is not null, it points to a place in memory containing an array of 50 elements.

That array is an array of chars, otherwise known as a string. You initialise it to an empty string but it's still there and the pointer to it (str) is still a valid, non-NULL pointer.

str!=NULL does not mean "str is not an empty string"! It means "str is not an undefined pointer".