Using lambda as an argument : std::function or tem

问题:

I'm studying c++11 especially interested in lambda.

After some practices, I assumed that lambda closure is an nameless function object.

So I wrote this code.

template <class callable_object>
void lambda_caller( callable_object lambda )
{
    std::cout<< sizeof(lambda) << endl;
    lambda();
}

I know that I can use std::function instead of using template, but I don't want the overhead while typecasting.

But I found one problem reading this question : Why can't I create a vector of lambda in C++11?

The answerer said, "Every lambda has a different type- even if they have the same signature.".

Compilers makes different codes for different classes.

So I think that my compiler will make another version of lambda_caller whenever I make another definition of lambda to pass.

Is there any way to avoid it, except using std::function? Isn't there any generic type for lambda closure?

回答1:

You can't avoid it. Lambda is just a class with operator()() overloaded which executes your code. So different code - different classes.

回答2:

std::function is the generic type for lambda closures. The problem is that each lambda may capture different variables. So it can't be reduced to say a function pointer and some data, because the lambda may have captured 3 variables or it may have captured 4. std::function will take care of making sure that enough memory is allocated for the data, but it comes at a cost(the data could be heap allocated).

However, if you are wanting to store several lambdas, and you know how many at compile-time. You could store them in an std::tuple instead. Which allows different types for each lambda. Unfortunately, C++ still doesn't provide a way to iterate over a tuple, but using Boost.Fusion you can.