Alias a template function with default parameters

Question

The following C++ code does not compile:

template <typename T>
void f(int, bool = true);

void g()
{
    auto h = f<int>;
    h(1); // error: too few arguments to function
}

Instead, I have to call h with the 2nd parameter:

h(1, true);

Why doesn't h(1) work?

Is there a simple way to alias a template function to bind the template argument while keeping the default function argument?

Answer 1

h is declared as function pointer, unfortunately it can't specify default arguments.

Default arguments are only allowed in the parameter lists of function declarations and lambda-expressions, (since C++11) and are not allowed in the declarations of pointers to functions, references to functions, or in typedef declarations.

You can use a lambda wrapping f instead. E.g.

auto h = [](int i) { f<int>(i); };
h(1); // -> f<int>(1, true), using f's default argument

Or specify default argument on the lambda too.

auto h = [](int i, bool b = true) { f<int>(i, b); };
h(1);        // -> f<int>(1, true), using h, i.e. lambda's default argument
h(1, true);  // -> f<int>(1, true), not using default argument
h(1, false); // -> f<int>(1, false), not using default argument