c++11 - 返回可修改引用的 minmax

Question

有了 C++ 的所有新特性(我认为 C++11 就足够了)，是什么阻止了 std::minmax 函数返回一对引用。

这样一来，如果一个人提供了两个可修改的引用，它们就可以被修改。这是在开 jar 头吗？

#include<functional>
// maybe all this options can be simplified
template<class T1, class T2> struct common;
template<class T> struct common<T, T>{using type = T;};
template<class T> struct common<T const&, T&>{using type = T const&;};
template<class T> struct common<T&, T const&>{using type = T const&;};
template<class T> struct common<T, T&>{using type = T const&;};
template<class T> struct common<T&, T>{using type = T const&;};
template<class T> struct common<T const&, T>{using type = T const&;};
template<class T> struct common<T, T const&>{using type = T const&;};

template<class T1, class T2, class Compare = std::less<>, class Ret = typename common<T1, T2>::type> 
std::pair<Ret, Ret> minmax(T1&& a, T2&& b, Compare comp = {}){
    return comp(b, a) ? 
        std::pair<Ret, Ret>(std::forward<T2>(b), std::forward<T1>(a))
        : std::pair<Ret, Ret>(std::forward<T1>(a), std::forward<T2>(b));
}

测试:

#include<cassert>
int main(){
    {
    int a = 1;
    int b = 10;
    auto& small = minmax(a, b).first;
    assert(small == 1);
    small += 1;
    assert(a == 2);
    }{
    int const a = 1;
    int b = 10;
    auto& small = minmax(a, b).first;
    assert(small == 1);
//    small += 1; error small is const reference, because a was const
    }{
    int a = 1;
    int const b = 10;
    auto& small = minmax(a, b).first;
    assert(small == 1);
//    small += 1; error small is const reference, because a was const
    }{
    int const a = 1;
    int const b = 10;
    auto& small = minmax(a, b).first;
    assert(small == 1);
//    small += 1; error small is const reference, because a was const
    }{
    int b = 10;
    auto& small = minmax(int(1), b).first;
    assert(small == 1);
//   small += 1; error small is const reference, because first argument was const
    }{
    int a = 1;
    auto& small = minmax(a, int(10)).first;
    assert(small == 1);
//   small += 1; error small is const reference, because second argument was const
    }
    {
    int const a = 1;
    auto& small = minmax(a, int(10)).first;
    assert(small == 1);
//    small += 1; error small is const reference, because both arguments are const
    }
    {
//    auto& small = minmax(int(1), int(10)).first; // error, not clear why
    auto const& small = minmax(int(1), int(10)).first; // ok
//    auto small2 = minmax(int(1), int(10)).first; // also ok
    assert(small == 1);
//    small += 1; error small is const reference, because both arguments are const
    }
}

Answer 1

很久以前，Howard Hinnant 有一篇类似这样的论文: N2199 .它非常开放的示例演示了您要解决的确切问题:

The function can not be used on the left hand side of an assignment:

int x = 1;
int y = 2;
std::min(x, y) = 3;  // x == 3 desired, currently compile time error

它接着列举了经常出现的悬空引用问题、混合类型以及对仅移动类型有用的示例，并继续提出了新版本的 min和 max解决了所有这些问题——它在底部包含了一个非常彻底的实现(太长了，无法粘贴到这里)。实现 minmax()基于此应该非常简单:

template <class T, class U,
    class R = typename min_max_return<T&&, U&&>::type>
inline
std::pair<R, R>    
minmax(T&& a, U&& b)
{
    if (b < a)
        return {std::forward<U>(b), std::forward<T>(a)};
    return {std::forward<T>(a), std::forward<U>(b)};
}

论文当时被拒了。它可能会回来。

能够取回可变引用固然不错，但能够避免悬空引用就更好了。匿名引用我最近看到的一个例子:

template<typename T> T sign(T); 

template <typename T> 
inline auto frob(T x, T y) -> decltype(std::max(sign(x - y), T(0))) { 
    return std::max(sign(x - y), T(0)); 
} 

This function has undefined behaviour for all inputs (the narrowest contract possible?).

请注意您的 common实现有这个问题。这些情况:

template<class T> struct common<T, T&>{using type = T const&;};
template<class T> struct common<T&, T>{using type = T const&;};
template<class T> struct common<T const&, T>{using type = T const&;};
template<class T> struct common<T, T const&>{using type = T const&;};

全挂了。如果我有，这意味着什么:

int i = 4;
auto result = your_minmax(i, 5);

result这是一个 pair<int const&, int const&> , 其中之一是对 i 的引用另一个悬垂着。所有这些情况都必须做 using type = T;为了安全。