Default lambda as templated parameter of a function

Question

Consider the following code

template void foo(const T& t = []() {}) {
  // implementation here
}

void bar() {
  foo([&         


        

Answer 1

Another (very efficient) way - default T to be a null functor.

// no_op is a function object which does nothing, regardless of how many
// arguments you give it. It will be elided completely unless you compile with
// -O0
struct no_op 
{ 
    template<class...Args>
    constexpr void operator()(Args&&...) const {} 
};

// foo defaults to using a default-constructed no_op as its function object
template<bool b, typename T = no_op> void foo(T&& t = T()) 
{    
  // implementation here
    t();
}

void bar() {
  foo<true>([&](){ std::cout << "something\n"; }); // this compiles
  foo<true>(); // this now compiles
}

Answer 2

You are trying to say something that makes no sense. You are asking the compiler to guess T from your arguments, but then you do not provide any argument.

The following code does compile, and does what you want:

template<bool b, typename T> void foo(const T& t) {
  // implementation here
}

template<bool b> void foo() {
  foo<b>([]() {}); // Call actual implementation with empty lambda
}

void bar() {
  foo<true>([&](){ /* implementation here */ }); // this compiles
  foo<true>(); // this now compiles as well
}

Answer 3

Consider overloading it directly:

template <bool b>
void foo(void) {
  foo([](){});
}

See CppReference:

Non-deduced contexts

4) A template parameter used in the parameter type of a function parameter that has a default argument that is being used in the call for which argument deduction is being done:

Type template parameter cannot be deduced from the type of a function default argument: template void f(T = 5, T = 7);

void g()
{
    f(1);     // OK: calls f<int>(1, 7)
    f();      // error: cannot deduce T
    f<int>(); // OK: calls f<int>(5, 7)
}

Answer 4

Default function arguments are not part of the template argument deduction process. To quote [temp.deduct.partial]/3:

The types used to determine the ordering depend on the context in which the partial ordering is done:

• In the context of a function call, the types used are those function parameter types for which the function call has arguments. ¹⁴¹

_{141) Default arguments are not considered to be arguments in this context; they only become arguments after a function has been selected.}

That bullet and note indicate that since you didn't provide an argument for t in the call to foo, the type T cannot be deduced. The default lambda argument can only be taken into account if the function is selected to be called, not before.

The solution, as all the others have noted, is to provide an overload without parameters, that will call the templated one with the default lambda you have in mind.

Answer 5

The compiler uses the arguments passed to deduce the template type. If there's no arguments, then how would the compiler be able to deduce the template type?

You can use overloading instead of default arguments here.

The overloaded non-argument function can simply call the function with the "default" argument:

template<bool b, typename T> void foo(const T& t) {
  // implementation here
}

template<bool b> void foo() {
  foo<b>([]() {});
}