The first one should probably be:
// amended first version
template<class T>
decltype(auto) max(T&& a, T&& b) {
return (a < b) ? std::forward<T>(b) : std::forward<T>(a);
}
Otherwise it wouldn't work for temporaries (the original first option fails with two temporaries).
But even now, with the new version above, it cannot work for a mix of an rvalue and an lvalue:
int i = my_max(3, 5); // ok
i = my_max(i, 15); // fails
// no known conversion from 'int' to 'int &&' for 1st argument
To add to that, the return value of the first option is either an lvalue-ref or an rvalue-ref (with or without const, depending on the arguments). Getting back rvalue-ref to a temporary would not be the best idea, it would work if we immediately copy from it, but then it would be better to return byvalue, which is not the approach taken by the first version.
The pitfall with the second, the const-lvalue-ref version, is that if a temporary is sent to it, we return a const-lvalue-ref to a temporary which is bug prone (well, not more than returning an rvalue-ref to a temporary, but still). If you copy it immediately you are fine, if you take it by-ref you are in the UB zone:
// second version
template<class T>
const T& max(const T& a, const T& b) {
return (a < b) ? b : a;
}
std::string themax1 = max("hello"s, "world"s); // ok
const std::string& themax2 = max("hello"s, "world"s); // dangling ref
To solve the above problem, with the cost of redundant copying for the lvalue-ref case, we can have another option, returning byvalue:
// third version
template<class T1, class T2>
auto max(T1&& a, T2&& b) {
return (a < b) ? std::forward<T2>(b) : std::forward<T1>(a);
}
The language itself took the 2nd option for std::max, i.e. getting and returning const-ref. And the user shall be careful enough not to take a reference to temporaries.
Another option might be to support both rvalue and lvalue in their own semantic, with two overloaded functions:
template<class T1, class T2>
auto my_max(T1&& a, T2&& b) {
return (a < b) ? std::forward<T2>(b) : std::forward<T1>(a);
}
template<class T>
const T& my_max(const T& a, const T& b) {
return (a < b) ? b : a;
}
With this approach, you can always get the result as a const-ref: if you went to the first one you get back a value and extend its lifetime, if you went to the second one you bind a const-ref to a const-ref:
int i = my_max(3, 5); // first, copying
const int& i2 = my_max(i, 25); // first, life time is extended
const std::string& s = my_max("hi"s, "hello"s); // first, life time is extended
const std::string s2 = my_max("hi"s, "hello"s); // first, copying
const std::string& s3 = my_max(s, s2); // second, actual ref, no life time extension
The problem with the above suggestion is that it only takes you to the lvalue-ref version if both arguments are const lvalue-ref. If you want to cover all cases you will have to actually cover them all, as in the code below:
// handle rvalue-refs
template<class T1, class T2>
auto my_max(T1&& a, T2&& b) {
return (a < b) ? std::forward<T2>(b) : std::forward<T1>(a);
}
// handle all lvalue-ref combinations
template<class T>
const T& my_max(const T& a, const T& b) {
return (a < b) ? b : a;
}
template<class T>
const T& my_max(T& a, const T& b) {
return (a < b) ? b : a;
}
template<class T>
const T& my_max(const T& a, T& b) {
return (a < b) ? b : a;
}
template<class T>
const T& my_max(T& a, T& b) {
return (a < b) ? b : a;
}
A last approach to achieve the overloading, and supporting all kind of lvalue-ref (const and non-const, including a mixture), without the need for implementing the 4 combinations for lvalue-ref, would be based on SFINAE (here presented with C++20 with a constraint, using requires):
template<class T1, class T2>
auto my_max(T1&& a, T2&& b) {
return (a < b) ? std::forward<T2>(b) : std::forward<T1>(a);
}
template<class T1, class T2>
requires std::is_lvalue_reference_v<T1> &&
std::is_lvalue_reference_v<T2> &&
std::is_same_v<std::remove_cvref_t<T1>, std::remove_cvref_t<T2>>
auto& my_max(T1&& a, T2&& b) {
return (a < b) ? b : a;
}
And if you bear with me for one past the last... (hope you are not in a bucket for monsieur state by now). We can achieve it all in one function!, and as a side benefit, even allow cases of comparison between derived and base if supported, so the following would work fine:
A a = 1;
B b = 2; // B is derived from A
// if comparison between A and B is supported, you can do
const A& max1 = my_max(a, b); // const-ref to b
const A& max2 = my_max(a, B{-1}); // const-ref to a temporary copied from a
const A& max3 = my_max(a, B{3}); // const-ref to B{3}
This would be the code to support this with a single function:
template<typename T1, typename T2>
struct common_return {
using type = std::common_reference_t<T1, T2>;
};
template<typename T1, typename T2>
requires std::is_lvalue_reference_v<T1> &&
std::is_lvalue_reference_v<T2> &&
has_common_base<T1, T2> // see code in link below
struct common_return<T1, T2> {
using type = const std::common_reference_t<T1, T2>&;
};
template<typename T1, typename T2>
using common_return_t = typename common_return<T1, T2>::type;
template<class T1, class T2>
common_return_t<T1, T2> my_max(T1&& a, T2&& b)
{
if(a < b) {
return std::forward<T2>(b);
}
return std::forward<T1>(a);
}
The machinery for the above can be found here.
For the variadic version of this, you can follow this SO post.
