Question

For the example

template <typename T>
void function(T&& arg)

Can somebody explain in detail how does it end that function signature becomes T& for lvalues and T&& for rvalues passed in ? I know that somehow(standard line needed) T -> T& in the case of lvalues and T -> T in case of revalues and then by combining & and && it results lvalue/rvalue reference.

Answer 1

The rule is found in section 8.3.2p6.

If a typedef, a type template-parameter, or a decltype-specifier denotes a type TR
that is a reference to a type T, an attempt to create the type “lvalue reference to cv TR” creates the type “lvalue reference to T“, while an attempt to create the type “rvalue reference to cv TR” creates the type TR.

Or in tabular form:

TR   R

T&   &  -> T&  // lvalue reference to cv TR -> lvalue reference to T
T&   && -> T&  // rvalue reference to cv TR -> TR (lvalue reference to T)
T&&  &  -> T&  // lvalue reference to cv TR -> lvalue reference to T
T&&  && -> T&& // rvalue reference to cv TR -> TR (rvalue reference to T)