std::function::target when T is a captureless lambda

Question

Consider the following code:

#include <functional>
#include <iostream>

void normalFunction()
{
    printf("bar");
}

int main()
{
    std::function<void ()> f1 = []()
    {
        printf("foo123");
    };

    std::function<void()> f2 = normalFunction;

    std::cout << f1.target<void (*)()>() << std::endl;
    std::cout << f2.target<void (*)()>() << std::endl;

    return 0;
}

The second line prints a valid address; the first call to target, however, fails (nullptr is returned), because the type of the stored callable is not void (), it's actually class <lambda_b356045a9ea53f6e7d502dd6abd8cae2> (as target_type tells us). This means that std::function<T>::target is very orthodox when checking the types and requires an exact match - even though the lambda can be converted to void (*)() without any problem, since it is captureless.

Is there a way to overcome this behavior? I want to be able to pass callbacks to a function that interfaces with C code, so it needs to pass function pointers further. One option is to just have that function take a function pointer, since lambdas without capture can be converted implicitly; but out of curiosity, I wonder if there's a way to do this using std::function.

std::function::target when T is a captureless lambda

0 Answers0