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 capture any entity.
[ Example:

void f2() {
    int i = 1;
    void g1(int = ([i]{ return i; })()); // ill-formed
    void g2(int = ([i]{ return 0; })()); // ill-formed
    void g3(int = ([=]{ return i; })()); // ill-formed
    void g4(int = ([=]{ return 0; })()); // OK
    void g5(int = ([]{ return sizeof i; })()); // OK
}

—end example ]

However, can we also use a lambda-expression itself as the default value for a function argument?

e.g.

template<typename functor>
void foo(functor const& f = [](int x){ return x; })
{
}

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);