获取 lambda 的 lambda 调用运算符的地址

Question

Is it possible to get the address of the lambda call operator for a lambda having captures? And even more: to assign such a pointer to a common pointer which can point to different lambdas having the same captures and the same calling-parameters? Strictly speaking, assigning a result (class1::*)(... ) to another result (class2::*)(... ) shouldn't be directly possible, but you might do dirty tricks and cast the pointers which could be possible since the lambda-classes are notionally interchangeable.

So what's the "proper" syntax for this? Even as a dirty trick.

The last line of the following code unfortunately doesn't work:

int main()
{
    int i;
    auto l100 = [&]()
    {
        i += 100;
    };
    using l100_t = decltype(l100);
    using l100_fn = void (l100_t::*)();
    l100_fn fn100 = &l100_t::operator ();
}

Answer 1

which can point to different lambdas having the same captures and the same calling-parameters

If you have some interface which expects the user to provide a "lambda" that "captures" a specific set of values, and takes a specific set of parameters, then it's not a lambda anymore. It's just a function pointer that is given a const& to a struct containing the "captured" values as one of its parameters. After all, the receiving code needs to store the "captured" values, right? So the user needs to provide a "capture" struct and a function pointer that takes the "capture" struct and the other arguments.

That's how you should build your API: clearly and explicitly. It's not the user's job to make sure they're capturing the right parameters; it's the API's job to tell the user want values are being "captured", and the user provides a function pointer that interfaces with them.

Yes, the user doesn't get to automatically treat those "captured" values as accessible by name; they have to use the parameter name to get at them. But the API becomes a lot more coherent and a lot less fragile.