c – 取消引用void指针时的reinterpret_cast行为

#include <iostream>

int main()
{
    int i{ 5 };
    void* v = &i;
    std::cout << reinterpret_cast<int&>(*v) << std::endl;
}

Live demo.GCC和VC都因我预期的错误而失败,抱怨代码试图在reinterpret_cast中取消引用void *.所以我决定在标准中查看这个.来自N3797,§5.2.10/ 11 [expr.reinterpret.cast]

A glvalue expression of type T1 can be cast to the type “reference to T2” if an expression of type “pointer to T1” can be explicitly converted to the type “pointer to T2” using a reinterpret_cast. The result refers to the same object as the source glvalue,but with the specified type. [ Note: That is,for lvalues,a reference cast reinterpret_cast<T&>(x) has the same effect as the conversion *reinterpret_cast<T*>(&x) with the built-in & and * operators (and similarly for reinterpret_cast<T&&>(x)). —end note ] No temporary is created,no copy is made,and constructors (12.1) or conversion functions (12.3) are not called.

在这种情况下,T1为void,T2为int,并且可以使用reinterpret_cast将void *转换为int *.因此满足所有要求.

根据说明,reinterpret_cast< int&>(* v)与* reinterpret_cast< int *>(&(* v))具有相同的效果,根据我的推算,它与* reinterpret_cast< int相同* GT;(v)中. 那么这是一个海湾合作委员会和VC的错误,还是铿锵有力,我错误地解释了这个？