Understanding template argument output with rvalue / lvalue

Question

This description from the function template does not recognize the value lvalue

Allows you to play with the following code:

#include <iostream> template <class T> void func(T&&) { std::cout<<"in rvalue\n"; } template <class T> void func(const T&) { std::cout<<"in lvalue\n"; } int main() { double n=3; func<double>(n); func(n); } 

He prints:

 in lvalue in rvalue 

I do not understand what is happening in the second call. How does the compiler resolve a template parameter? Why is there no ambiguity?

Answer 1

When you say func<double>(n) , there are no arguments, because you specify an argument, and so the choice is between func(double &&) and func(const double &) . The first is not viable since the rvalue reference cannot communicate with the lvalue (namely n ).

Only func(n) performs the output of the argument. This is a complex topic, but in a nutshell you have two possible candidates:

 T = double &: func(T &&) --> func(double &) (first overload) T = double: func(const T &) --> func(const double &) (second overload) 

The first overload is strictly better, because it requires less conversion of the argument value (namely, from double to const double ).

A magic ingredient is a link that compresses " , which means that T && can be an lvalue reference when T itself a reference type (especially double & && becomes double & , and this allows the first conclusion).