I want to default a lambda argument in a template function but it fails to compile. What am I missing?
template <typename F>
void foo(
F f = [](){
std::cout << "Hello!\n";
}
) {
f();
}
int main() {
foo(); // 1. does not compile
foo([](){ // 2. this is ok
std::cout << "Hello!\n";
});
}
You can't deduce the template parameter of a function from the default function arguments. See this question for details on why this restriction is in place.
So you must provide a default template parameter yourself. Since you need both the type and the value of the lambda, a simple way to do this would be to write the lambda once, and then use it inside the function template.
auto lambda = []()
{
std::cout << "Bye!\n";
};
template <typename F = decltype(lambda)> // default parameter
void foo(F f = lambda) // default value
{
f();
}
Here's a demo
