问题描述
如果我写
int zero = 0;
void *p1 = (void *)0;
void *p2 = (void *)(int)0;
void *p3 = (void *)(0 /*no-op, but does it affect the next zero?*/, 0);
void *p4 = (void *)zero; // For reference, this is a pointer to address zero
void *p5 = 0; // For reference, this is a null pointer
void *p6 = NULL; // For reference, this is a null pointer
void *p7 = nullptr; // For reference, this is a null pointer (C++11)
static const int static_zero_1 = 0; // Is this a literal zero when used?
static const int static_zero_2 = 1 - 1; // No "literals 0" per se... is it?
void *p8 = (void *)static_zero_1; // I have seen weird substitution rules...
void *p9 = (void *)static_zero_2; // do they apply for NULL too?
p1、p2 和 p3 中的哪一个(我添加了 p8 和 p9) 将是 空指针(即 == NULL，可能是也可能不是地址零)，它们中的哪些是是地址为零的指针(可能是也可能不是 ==NULL)?
which of p1, p2, and p3 (edit: I added p8 and p9) would be null pointers (i.e. == NULL, may or may not be address zero), and which of them would be pointers with the address zero (may or may not be == NULL)?
如果答案在 C 和 C++ 中不同，那么它们分别是什么?
If the answer is different in C and C++, what is it in each of them?
推荐答案
p1 和 p2 为空指针；p3 是实现定义的，并且可能是别的东西.(逗号运算符不能是一个常量表达式.和一个非常量的映射指针的整数值 0 是实现定义的.)C 是此处与 C++ 相同.
p1 and p2 are null pointers; p3 is implementation defined,
and may be something else. (A comma operator cannot be part of
a constant expression. And the mapping of a non-constant
integral value 0 to a pointer is implementation defined.) C is
identical to C++ here.
p8 和 p9 在 C++ 中都是空指针，但在 C 中不是.
p8 and p9 are both null pointers in C++, but not in C.
关于您对 static_zero_2 的评论，没有任何一种语言都要求存在文字零，任何地方.g++ 将 NULL 定义为编译器内置的 __null，例如，您可以使用 (1 - 1) 或 ' ' 或任何其他计算结果为 0 的常量表达式.
With regards to your comment on static_zero_2, there is no
requirement in either language that a literal zero be present,
anywhere. g++ defines NULL as the compiler built-in __null,
for example, and you can use (1 - 1), or ' ', or any other
constant expression evaluating to 0.
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