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?
这 P1 ， P2 ，而 P3 （修改我加入 P8 和 P9 ）会的空指针的（即 == NULL ，可能会或可能不会是零地址），其中这将与地址指针为零（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)?
如果答案是C和C ++不同，它是什么在他们每个人的？
If the answer is different in C and C++, what is it in each of them?
推荐答案
P1 和 P2 是空指针; P3 是实现定义，
并可能是别的东西。 （逗号操作者不能成为其中的一部分
恒定的前pression。和非恒定的映射
整数值0到指针实现定义。）C是
与C ++在这里。
p1 and p2 are null pointers; p3 is implementation defined,and may be something else. (A comma operator cannot be part ofa constant expression. And the mapping of a non-constantintegral value 0 to a pointer is implementation defined.) C isidentical to C++ here.
P8 和 P9 在C两个空指针++，但不是在C
p8 and p9 are both null pointers in C++, but not in C.
至于您的评论 static_zero_2 ，也没有
在任何语言要求的字面零是present，
任何地方。 G ++定义 NULL 因为编译器内置的 __空，
例如，你可以使用（1 - 1）或'\\ 0'，或其他任何
恒恩pression评估为0。
With regards to your comment on static_zero_2, there is norequirement 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 '\0', or any otherconstant expression evaluating to 0.
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