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
void foo(T = 0);

then foo(0); will call foo but foo() is ill-formed. You'd need to call foo() 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);