Type deduction for template references

Question

Let's consider the following code:

template<typename T>
void f(T&);

int x=0;
int y=1;
f(x+y);

It seems that type deduction happens for this code. But why? There is no reference collapsing rule that will produce int&& at the end.

Answer 1

According to the C++17 standard, [temp.deduct.call]/3:

... If P is a reference type, the type referred to by P is used for type deduction. ...

Here, P is T&, so the compiler removes the reference and compares T with the type of the argument x+y. It therefore deduces T as int. Then, at the overload resolution stage, an error occurs because an rvalue of type int cannot be bound to the parameter of type int&.

Template argument deduction must occur before overload resolution, because otherwise the compiler would not know what parameter types to use for overload resolution.