是否可以安全地转换为具有相同数据成员布局但具有不同实现的类？ [英] Is it safe to cast to a class that has the same data member layout, but a different implementation?

问题描述

第一个类将用于私有继承，以确保完全相同的布局。这将使投射更安全。

The first class will be used for private inheritance in order to ensure the exact same layout. This should make casting safe.

#include <iostream>
#include <string>

struct data_base
{
    data_base( int i, std::string&& s ) noexcept
        : i_{ i }
        , s_{ std::move( s ) }
    {}

    int i_;
    std::string s_;
};

在这个简单的例子中，我打印 int 数据成员，随后是 data 实例的 std :: string 数据成员。 >

In this trivial example, I print the int data member first followed by the std::string data member for instances of data<true>.

template<bool = true>
struct data : private data_base // inherits
{
    data( int i, std::string&& s ) noexcept
        : data_base( i, std::move( s ) )
    {}

    void print()
    {
        std::cout << "data<true> - " << i_ << s_ << '\n';
    }
};

但是， data< false> std :: string 数据成员，随后是 int 数据成员。

However, the data<false> prints the std::string data member first, followed by the int data member.

template<>
struct data<false> : private data_base
{
    void print()
    {
        std::cout << "data<false> - " << s_ << i_ << '\n';
    }
};

示例：

int main()
{
    data<true> d{ 5, "abc" };
    d.print();
    ( ( data<false>& )d ).print();
}

演示： http://coliru.stacked-crooked.com/a/8b1262afe23dc0a2

演示显示，即使使用 -fstrict-aliasing 标志，没有警告。

As the demo shows, even with the -fstrict-aliasing flag on, there's no warnings.

现在，由于他们有相同的布局，我可以在两种类型之间转换，以获得一种不同类型的静态多态性;

Now, since they have the same layout, I thought that I could just cast between the two types in order to get a different kind of static polymorphism; without the cost of a virtual function call.

此使用是否安全或是否触发未定义的行为？

推荐答案

它或多或少描述了这里，所谓的 boost突变成语

It's more or less what's described here, the so called boost mutant idiom.

：

Boost突变成语使用 reinterpret_cast ，并且在很大程度上取决于具有相同数据成员（类型和顺序）的两个不同结构的存储器布局是可互换的。 虽然C ++标准不保证此属性，但实际上所有编译器都满足。此外，如果仅使用POD类型，则突变成语是标准的。

Boost mutant idiom makes use of reinterpret_cast and depends heavily on assumption that the memory layouts of two different structures with identical data members (types and order) are interchangeable. Although the C++ standard does not guarantee this property, virtually all the compilers satisfy it. Moreover, the mutant idiom is standard if only POD types are used.

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注意：该页面已过时，我不知道最近的修订版是否改变了上述保证金。

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Note: that page is pretty outdated, I don't know if the most recent revisions changed something about the guarantees above mentioned.

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