minmax that returns modifiable references - c++11

Question

With all the new features of C++ (C++11 suffices I think), what prevents from having a std::minmax function that returns a pair of references.
In this way, if one feeds two modifable references, they can be modified. Is this opening a can of worms?
#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));
}
Test:
#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

There was a paper kind of along these lines a long time ago, by Howard Hinnant: N2199. Its very opening example demonstrates the precise problem you're trying to solve:
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
It goes on to list as examples the frequently dangling reference problem, mixing types, and being useful with move-only types, and goes on to propose new versions of min and max that address all of these problems - it includes a very thorough implementation at the bottom (which is too long to paste here). Implementing minmax() based on that should be pretty straightforward:
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)};
}
The paper was rejected at the time. It's possible it could come back though.
Being able to get back mutable references is nice, but being able to avoid dangling references is even nicer. Anonymously quoting from an example I saw recently:
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?).
Note that your common implementation has this problem. These cases:
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&;};
all dangle. What this means if I have:
int i = 4;
auto result = your_minmax(i, 5);
result here is a pair<int const&, int const&>, one of which is a reference to i and the other of which dangles. All of these cases have to do using type = T; in order to be safe.