Question

使用C ++的所有新功能（我认为C ++ 11就足够了），可以防止使用std::minmax函数返回一对引用。

通过这种方式，如果一个提供两个可修改的引用，则可以修改它们。这个开口是一堆蠕虫吗？

#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：N2199就曾有过类似的论文。其开头的示例演示了您要解决的确切问题：

该功能不能在作业的左侧使用：
int x = 1;
int y = 2;
std::min(x, y) = 3;  // x == 3 desired, currently compile time error

接着列举了频繁出现的参考问题，混合类型以及对仅移动类型有用的示例，并继续提出了解决所有问题的min和max新版本这些问题-它在底部包括一个非常全面的实现（太长了，无法在此处粘贴）。基于此实现minmax()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)};
}

当时论文被拒绝了。可能它会回来。

能够取回可变引用很不错，但是能够避免 dangling 引用则更好。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)); 
} 
此函数对所有输入均具有不确定的行为（最窄合同可能吗？）。

请注意，您的common实现存在此问题。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&;};

所有悬挂。如果我有这意味着什么？

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

result这里是pair<int const&, int const&>，其中一个是对i的引用，另一个是悬挂的。为了安全起见，所有这些情况都必须执行using type = T;。

minmax返回可修改的引用

1 个答案: