在将指针转换为uintptr_t之前需要转换为void *吗?反之亦然? [英] Is a conversion to `void *` required before converting a pointer to `uintptr_t` and vice versa?

问题描述

C99标准的 7.18.1.4 部分:

以下类型指定了一个无符号整数类型，其属性是可以将指向 void 的任何有效指针转换为该类型，然后再转换回指向的指针 void ，结果将等于原始指针:

The following type designates an unsigned integer type with the property that any valid pointer to void can be converted to this type, then converted back to pointer to void, and the result will compare equal to the original pointer:

uintptr_t

uintptr_t

这是否意味着只能将 void * 类型转换为 uintptr_t 并返回，而无需更改原始指针的值?

Does this mean that only void * types can be converted to uintptr_t and back without changing the value of the original pointer?

特别是，我想知道是否需要以下代码才能使用 uintptr_t :

In particular, I would like to know if the following code is required to use uintptr_t:

int foo = 42;
void * bar = &foo;
uintptr_t baz = bar;
void * qux = baz;
int quux = *(int *)qux; /* quux == foo == 42 */

或者如果C99标准保证此更简单的版本产生相同的效果:

Or if this simpler version is guaranteed by the C99 standard to result in the same effect:

int foo = 42;
uintptr_t bar = &foo;
int baz = *(int *)bar; /* baz == foo == 42 */

在将指针转换为 uintptr_t 以及反之亦然之前是否需要转换为 void * ?

Is a conversion to void * required before converting a pointer to uintptr_t and vice versa?

推荐答案

之所以存在该区别，是因为尽管指向 object 的任何指针都可以转换为 void * ，C不需要将函数指针可以转换为 void * 并再次返回！

The distinction exists also because while any pointer to an object can be converted to void *, C doesn't require that function pointers can be converted to void * and back again!

对于指向其他类型对象的指针，C标准表示 any 指针可以转换为整数，并且整数可以转换为任何指针，但是这种结果是实现定义的.由于标准规定只有 void * 可来回转换，所以最安全的选择是首先 cast 指向 void * 的指针，因为它可能是当转换为 uintptr_t 时具有不同表示形式的指针也将导致不同的整数表示形式，因此可以想到:

As for pointers to objects of other types, the C standard says that any pointer can be converted to an integer, and an integer can be converted to any pointer, but such results are implementation-defined. Since the standard says that only void * is convertible back and forth, the safest bet is to cast the pointer to void * first, as it might be that pointers that have different representations when converted to uintptr_t will result in a different integer representation too, so it could be conceivable that:

int a;
uintptr_t up = (uintptr_t)&a;
void *p = (void *)up;
p == &a; // could be conceivably false, or it might be even that the value is indeterminate.

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