c++ - 比较 uintptr_t 和指针类型的最佳方法是什么？

Question

它只是 reinterpret_cast 吗？

int *pointer;
uintptr_t value;
value == reinterpret_cast<uintptr_t>(pointer);

Answer 1

真的取决于你的目标。

[expr.reinterpret.cast]

4 A pointer can be explicitly converted to any integral type large enough to hold it. The mapping function is implementation-defined. [ Note: It is intended to be unsurprising to those who know the addressing structure of the underlying machine.  — end note ] A value of type std​::​nullptr_­t can be converted to an integral type; the conversion has the same meaning and validity as a conversion of (void*)0 to the integral type.

5 A value of integral type or enumeration type can be explicitly converted to a pointer. A pointer converted to an integer of sufficient size (if any such exists on the implementation) and back to the same pointer type will have its original value; mappings between pointers and integers are otherwise implementation-defined.

映射是实现定义的(很明显)。如果你想检查 pointer 的值用于初始化 value ，那么你的支票是不够的。以上并不保证 reinterpret_cast<uintptr_t>(pointer)将始终产生相同的整数，即使今天所有理智的实现都是如此。

我会反向检查，因为我们有往返保证:

reinterpret_cast<int*>(value) == pointer;

但即便如此，它的保证也很薄弱。如果我是你，我不会过多地关注这些转换。可能值得重新考虑您的设计。