Instantiating C++ lambda by its type

Question

I want a way to make functor from function. Now I trying to wrap function call by lambda function and instantiate it later. But compiler says than lambda constructor is dele

Answer 1

The code doesn't make sense. Imagine you have a capturing lambda like this:

{
    int n = 0;
    auto t = [&n](int a) -> int { return n += a; };
}

What could it possibly mean to default-construct an object of type decltype(t)?

As @Matthieu suggests, you could wrap the lambda into a function object:

std::function F = t;

Or you could template your call-site directly on the type of the lambda (or any callable entity):

template 
int compute(int a, int b, F f)
{
    return a * f(b);  // example
}

Usage: int a = 0; for (int i : { 1, 3, 5 }) { a += compute(10, i, t); }

If at all possible, the second style is preferable, since the conversion to std::function is a non-trivial, potentially expensive operation, as is the actual function call through the resulting object. However, if you need to store a uniform collection of heterogeneous callable entities, then std::function may well be the easiest and most convenient solution.