Lambda black magic taking different types

user_806286 0
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Trying to find yet another way to simply serialize my code, I had the stupid idea to try this impossible thing. But it worked. I really wonder why:

template <typename C>
void f(C c)
{
    int a = 1;
    float b = 1.5f;
    c(a);
    c(b);
}

int main()
{
    f([](auto v){
        std::cerr << v << "\n";
    });
    return 0;
}

I looked at the resulting assembly (g++-9.3 -O0 -g -S -fverbose-asm test.cpp -o test.s), and it appears that two distincts lambdas are generated: one taking a float, and the other one taking an int. Is this black magic? Is there anyone who knows the standard and can explain more about this?

auto lambda c++
2 Answers

A generic lambda is mostly similar to

struct lambda
{
    template <typename T>
    auto operator()(T v) const
    {
        std::cerr << v << "\n";
    }
// ...
};

The lambda itself is not a template, but its member operator() is.

user_26845390

This call:

f([](auto v){
        std::cerr << v << "\n";
    });

is passing a generic lambda to f, as the argument c. A generic lambda is one that has a templated member operator(), because of the auto parameter.

So when you make the following calls:

c(a);  // int argument
c(b);  // float argument

the compiler will instantiate one version of the lambda's member operator() with int, and one version of the member operator() with float.

user_83728530
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