为什么 assert(str?=NULL) 不返回错误

Question

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);
}

Answer 1

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).

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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 .

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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.

Answer 2

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"