c - 为什么在 C 的 if-else 条件中传递 (!NULL) 为真？

Question

在这几行代码中，

void main()
{
   if(!NULL)
     {
       printf("one.");
     }
   else
     {
       printf("two");
     }
}

输出是一个。这是为什么？为什么 !NULL 为真？

Answer 1

来自 C 标准#6.3.2.3p3 [已添加强调]

3 An integer constant expression with the value 0, or such an expression cast to type void *, is called a null pointer constant.⁶⁶⁾ If a null pointer constant 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.

....
....

⁶⁶⁾ The macro NULL is defined in <stddef.h> (and other headers) as a null pointer constant; see 7.19.

来自 C 标准#6.5.3.3p5

5 The result of the logical negation operator ! is 0 if the value of its operand compares unequal to 0, 1 if the value of its operand compares equal to 0. The result has type int. The expression !E is equivalent to (0==E).

所以，这个声明

if(!NULL)

相当于

if (0==NULL)

0==NULL被评估为 true .因此，您得到输出 one .

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附加:

void main() 的返回类型不符合标准。相反，您应该使用 int作为 main() 的返回类型.