The expresssion (void *)0 is called a null pointer.
But how about the following:
int i = 0;
void *s = (void *)i;
Is s also a null-pointer? The C-language standard says:
6.3.2.3 Pointers
3 An integer constant expression with the value 0, such an expression cast to type void *, or the predefined constant nullptr is called a null pointer constant70). If a null pointer constant or a value of the type nullptr_t (which is necessarily the value nullptr) is converted to a pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal to a pointer to any object or function.
4 Conversion of a null pointer to another pointer type yields a null
pointer of that type. Any two null pointers shall compare equal.5 An integer may be converted to any pointer type. Except as previously specified, the result is implementation-defined, might not be correctly aligned, might not point to an entity of the referenced type, and might produce an indeterminate representation when stored into an object.71)
According to this s would not be a null pointer?
5
Yes, s is a null-pointer. Because (void*)i equal to (void *)0 which is a null-pointer. You can check it like this:
gsm@fc:~/tmp$ cat a.c
#include <stdio.h>
int main()
{
int i = 0;
void* s = (void*)i;
printf("s %s null-pointern", s==NULL ? "is" : "isn't");
return 0;
}
gsm@fc:~/tmp$ gcc -std=c89 a.c
a.c: 在函数‘main’中:
a.c:6:15: 警告：将一个整数转换为大小不同的指针 [-Wint-to-pointer-cast]
6 | void* s = (void*)i;
| ^
gsm@fc:~/tmp$ ./a.out
s is null-pointer
2