Can we use a lambda-expression as the default value for a function argument?

Question

Refering to the C++11 specification (5.1.2.13):

A lambda-expression appearing in a default argument shall not implicitly or explicitly capt

Answer 1

Yes. In this respect lambda expressions are no different from other expressions (like, say, 0). But note that deduction is not used with defaulted parameters. In other words, if you declare

template<typename T>
void foo(T = 0);

then foo(0); will call foo<int> but foo() is ill-formed. You'd need to call foo<int>() explicitly. Since in your case you're using a lambda expression nobody can call foo since the type of the expression (at the site of the default parameter) is unique. However you can do:

// perhaps hide in a detail namespace or some such
auto default_parameter = [](int x) { return x; };

template<
    typename Functor = decltype(default_parameter)
>
void foo(Functor f = default_parameter);