用C类型转换NULL指针 [英] Typecasting NULL pointer in C

问题描述

我有一个结构

struct detail {
int id;
uintptr_t init;
// blah blah
};
struct detail info;
info.id = 1;
info.init = (uintptr_t)NULL;

我要做出的的init 成员NULL。可能/可能不会发生什么，如果我类型转换（或不强制转换） NULL ？如果我直接给它NULL像 info.init = NULL; 这会让相对于运行时错误任何区别。它编译罚款。但是，code的执行是我最关心的。

I have to make the the init member NULL. What may/may not happen if I typecast(or do not typecast) NULL ? What if I directly assign it NULL like info.init = NULL; Does it make any difference with respect to runtime errors. It compiles fine. But the execution of the code is my main concern.

感谢

推荐答案

有在标准不能保证，如果 PTR 是一个空指针，则（uintptr_t形式）PTR 是 0 。

There is no guarantee in the standard that if ptr is a null pointer, then (uintptr_t)ptr is 0.

如果你不关心哪个空指针和零整数是不等价的系统，则 info.init = 0; 的罚款。

If you don't care about systems on which null pointers and zero integers aren't equivalent, then info.init = 0; is fine.

的的init 成员都有整数类型，它不能是造空。您可以指定 0 它，或者你可以给它分配一个空指针转换为 uintptr_t形式的结果。在几乎每一个C实现过，这些都是同样的事情。但它不能保证，已经有系统的，它是不一样的。

The init member has integer type, it cannot be "made null". You can assign 0 to it, or you can assign to it the result of converting a null pointer to uintptr_t. On almost every C implementation ever, those are the same thing. But it is not guaranteed, and there have been systems on which it is not the same.

NULL 可能是一个空指针，或者它可能是一个整型常量 0 。在后一种情况下，是的标准保证，（uintptr_t形式）（NULL）是 0 。因此，有可能实现在其 info.init = NULL; （无效*）（info.init）; 是未定义行为。它不会导致空指针的整数相当于空是不为0，并计算了无效的指针的值是UB。

NULL might be a null pointer, or it might be an integer constant 0. In the latter case, there is a guarantee in the standard that (uintptr_t)(NULL) is 0. So there can be implementations on which info.init = NULL; (void*)(info.init); has undefined behavior. It wouldn't result in a null pointer if the integer equivalent of null isn't 0, and computing an invalid pointer value is UB.

所以，如果你想保证信息，当转换为指针类型，结果在一个空指针那么对于真正的便携性，你应该做的 info.init =（uintptr_t形式）（无效*）（NULL）; 。你可以任意给读者一个额外的线索，通过包括在 uintptr_t形式将被转换为指针类型，而不是无效* 。很少有很好的理由，用于存储 uintptr_t形式，因此提示是怎么回事也许读者有所帮助。

So, if you want to guarantee that info, when converted to a pointer type, results in a null pointer then for true portability you should do info.init = (uintptr_t)(void*)(NULL);. You could optionally give the reader an extra clue by including the pointer type that the uintptr_t is going to be converted to, instead of void*. There are very few good reasons for storing a uintptr_t, so hints what is going on might help the reader.

请注意，有在零值恒前pression，转换为指针类型标准的保证，是一个空指针。这确实的不的暗示零值不恒定的前pression，转换为指针类型，是一个空指针。它也不意味着空指针，转换为整数类型，0那些过去的​​两件事情是在大多数实现真正的（包括所有现代的）。

Note that there is a guarantee in the standard that a zero-valued constant expression, converted to pointer type, is a null pointer. This does not imply that a zero-valued non-constant expression, converted to a pointer type, is a null pointer. Neither does it imply that a null pointer, converted to integer type, is 0. Those last two things happen to be true in most implementations (including all "modern" ones).

这篇关于用C类型转换NULL指针的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！