- A+
I was surprised to find this code compiles:
#include <functional> struct Callable { void operator() () { count++; } void operator() () const = delete; int count = 0; }; int main() { const Callable counter; // counter(); //error: use of deleted function 'void Callable::operator()() const' std::function<void(void)> f = counter; f(); const auto cf = f; cf(); } https://wandbox.org/permlink/FH3PoiYewklxmiXl
This will call the non-const call operator of Callable. Comparatively, if you do const auto cf = counter; cf(); then it errors out as expected. So, why does const correctness not seem to be followed with std::function?
std::function adds a layer of indirection, and this layer of indirection does not pass through constness to the callable.
I'm not really sure why this is — probably because std::function takes a copy of the callable and has no need to keep the copy const (in fact this might break assignment semantics) — I'm also not really sure why you'd need it to.
(Of course, directly invoking operator() on an object of a type that you just so happened to call Callable and declared as const will require a const context, as it would with any other object.)
Best practice is to give the callable a const operator() and leave it at that.
tl;dr: true, but not a bug, and doesn't matter
