A转换的C样式转换可以处理，但C ++转换不能 [英] A conversion that C-style cast can handle, but C++ casts cannot

问题描述

据说是C样式转换只是试图适用的C ++铸件不同组合和使用第一允许组合。不过，我有听说有只C样式转换可以处理的情况下，虽然没有C ++的类型转换的组合被允许的感觉。

It is said that C-style cast just tries to apply different combination of C++ casts and the first allowed combination is used. However, I have a feeling that I heard that there are situations that only C-style cast can handle, while none of combination of C++ casts are allowed.

难道我错了吗？那是真的，任何的C风格的铸造的任何的情况下（在C ++）能够与C ++类型转换的适当的组合来代替？

Am I wrong? Is that true that any C-style cast in any context (in C++) can be replaced with a proper combination of C++ casts?

UPD 感谢欢呼和心连心。 - 阿尔夫，我们有一个例子，C ++转换不能在它们不能产生的定义的和的有望的行为的意义处理。先进的问题是提供C ++转换不能处理意甚的编译它不可能是一个例子的？

UPD Thanks to Cheers and hth. - Alf, we have an example that C++ casts cannot handle in the meaning they cannot produce defined and expected behavior. Advanced question is to provide an example which C++ casts cannot handle meaning it cannot be even compiled?

推荐答案

投放到人迹罕至的基础只能是pssed为C风格的类型转换（语法变种之一）前$ P $。在这种情况下它相当于一个的static_cast ，这可能更改地址，但的static_cast 无法访问基地。

Cast to inaccessible base can only be expressed as a C style cast (one of the syntactic variants). In that context it is equivalent to a static_cast, which may change the address, except that static_cast can't access the base.

例如：

struct Base
{
    int x = 42;
};

struct Oh_my
    : private Base
{
    virtual ~Oh_my() {}
};

#include <iostream>
using namespace std;
auto main() -> int
{
    Oh_my o;
    cout << "C cast: " << ((Base&)o).x << endl;
    cout << "reinterpret_cast: " << reinterpret_cast<Base&>(o).x << endl;
}

在Windows 7中使用MinGW的g ++输出：

Output with MingW g++ in Windows 7:

C cast: 42
reinterpret_cast: 4935184

但因为它是pretty未定义行为，最后输出操作可能只是崩溃。

But since it's pretty Undefined Behavior, the last output operation could just crash.

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