将非`void`指针转换为`uintptr_t`，反之亦然

问题描述

有两个相关的C标准规则:

There are two related C standard rules:

C99标准，6.3.2.3:

和7.20.1.4:

这意味着符合以下代码:

It means, that the following code is compliant:

int *p = NULL;
void *q = (void*)p;
uintptr_t s = (uintptr_t)q;

但是它真的需要分两步进行吗?如果执行以下操作，编译器会执行隐式中间强制转换吗?

But does it really need the two-step cast? Will the compiler perform an implicit intermediate cast if doing something like:

int *p = NULL;
uintptr_t s = (uintptr_t)p;

(嗯，它可能适用于大多数编译器，但是我的问题是关于标准合规性的问题)

(Well, it probably will on most compilers, but my question is about standard compliance)

推荐答案

我不会冒险.该标准明确规定了允许和禁止的条件.

I wouldn't risk it. The standard makes it abundantly clear what is allowed and what is not allowed.

编写uintptr_t s = (uintptr_t)(void*)p;会向代码阅读器发出信号，表明您知道自己在做什么.

Writing uintptr_t s = (uintptr_t)(void*)p; signals to a reader of your code that you know what you're doing.

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