How to invoke a lambda template?

Question

I was able to compile the following code with gcc:

template<typename... Pack>
auto func(Pack... x) {
    return (x + ...) ;
}

template<typename... Pack>
auto lamd = [](Pack... x) {
    return (x + ...) ;
};

I can invoke the function template with func(1,2,3), but I get an error when invoking a the lambda, with lamd(1,2,3) or lamd<int>(1,2,3).

Answer 1

The second definition is a variable template. It does not define the lambda's operator() as a template, but rather takes the parameter pack for the argument types of operator(). The resulting operator() is a regular member function of the instantiated variable's closure type. There is no template argument deduction possible here.

So when you write lamd<int>, the variable gets a closure type with a operator()(int), not something callable with 3 integers.

As mentioned already, you can use a generic lambda instead.

In C++20, if you'd need the lambda's argument types to be named and deduced, you can use the syntax:

auto lamd = []<typename... Pack>(Pack...) {}

This will define the operator as template, accepting a parameter pack, and leave the door open for template argument deduction.