How does template argument deduction work when an overloaded function is involved as an argument?

Question

This is the more sophisticated question mentioned in How does overload resolution work when an argument is an overloaded function?

Below code compiles without any problem:

void foo() {}
void foo(int) {}
void foo(double) {}
void foo(int, double) {}

// Uncommenting below line break compilation
//template<class T> void foo(T) {}

template<class X, class Y> void bar(void (*f)(X, Y))
{
    f(X(), Y());
}

int main()
{
    bar(foo);
}

It doesn't appear a challenging task for template argument deduction - there is only one function foo() that accepts two arguments. However, uncommenting the template overload of foo() (which still has just a single parameter) breaks compilation for no obvious reason. Compilation fails both with gcc 5.x/6.x and clang 3.9.

Can it be explained by the rules of overload resolution/template argument deduction or it should be qualified as a defect in those compilers?

TartanLlama TartanLlama · Accepted Answer · 2016-11-01T09:36:47

As noted in the answer to your linked question:

[temp.deduct.call]/6:When P is a function type, pointer to function type, or pointer to member function type:

— If the argument is an overload set containing one or more function templates, the parameter is treated as a non-deduced context.

Since the overload set contains a function template, the parameter is treated as a non-deduced context. This causes template argument deduction to fail:

[temp.deduct.type]/4: [...]If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.

And this failed deduction gives you your error. Note that if you explicitly specify the arguments, the code compiles successfully:

bar<int,double>(foo);

Live demo

How does template argument deduction work when an overloaded function is involved as an argument?

1 Answers