Type Erasure and the Cost of a Callable
A reported C++ interview track asks about std::function alongside RAII and std::move, and it is a well-chosen question: answering it properly requires explaining type erasure, which is the technique behind half the standard library's flexibility and most of its hidden costs.
The problem type erasure solves
A lambda has a unique, unnameable type. Two lambdas taking an int and returning void are different types with nothing in common, so a container cannot hold both and a function cannot take either without being a template.
std::function<void(int)> is one type that can hold any of them. It does that by storing the callable behind an interface it synthesises: a pointer to the object, and a small table of operations, call, copy, destroy, that know the concrete type. That is type erasure. The type is not really erased, it is hidden behind a uniform interface generated per stored type.
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