C++11 Lambda functions implicit conversion to bool vs. std::function

Question

Consider this simple example code:

#include 
#include 

void f(bool _switch) {
    std::cout << \"Nothing really\" &l         


        

Answer 1

A lambda function with no capture can be converted to a regular function pointer, which then has a standard conversion to a bool.

If you take the std::function by non-const reference, then that eliminates it as a candidate, since converting the lambda to a std::function requires a temporary, and a temporary cannot bind to a non-const reference. That just leaves f(bool) as a candidate, so there is no ambiguity.

There are many ways you could avoid the ambiguity. For example, you could create a std::function variable first:

std::function g = [](int _idx){ return 7.9;};
f(g);

or you could cast the lambda:

f(std::function([](int _idx){return 7.9;}));

You could have a helper function:

template
std::function make_function(T *f) { return {f}; } 

int main ( int argc, char* argv[] ) {
    f(make_function([](int _idx){ return 7.9;}));
    return 0;
}  

or you could grab the particular function you are interested in:

int main ( int argc, char* argv[] ) {
    void (*f_func)(std::function) = f;
    f_func([](int _idx){ return 7.9;});
    return 0;
}