Emulate copy-assignment operator for lambdas in C++ - c++

Question

This question has two parts
Firstly, can someone explain the rationale behind C++ disabling the copy-assignment operator for lambdas? If you're going to allow the copy constructor, why not the copy-assignment operator?
Secondly, how do you best overcome this limitation without forcing people to write C++03 style functors, or using std::function (the functions I'm dealing with are tiny, and I'd like the compiler to inline them wherever possible)?
Background:
I'm trying to implement a flat_map like operation in a stream processing library I'm writing, similar to flatMap in Scala or other functional languages. As a result, I need to create an iterator that iterates over a list of iterators. Each time the flat_map iterator is de-referenced a lambda associated with the inner iterator is executed. The outer iterator needs to switch the inner iterator each time the inner iterator reaches the end. Since the inner iterator contains a lambda, and therefore does not have a copy-assignment operator, it's not possible to switch it. Technically I could solve the problem using dynamic allocation, so that I always call the copy-constructor, but that doesn't seem like the right approach. Here is a snippet of code that might help highlight the problem:
template <typename Iter>
class flat_map_iterator {
public:
flat_map_iterator& operator++() {
++it_inner_;
if (it_inner_ == (*it_outer_).end()) {
++it_outer_;
// ERROR: cannot be assigned because its copy assignment operator is implicitly deleted
it_inner_ = (*it_outer_).begin();
}
return *this;
}
private:
Iter it_outer_;
typename Iter::value_type::iterator it_inner_;
};
Edit:
Thanks for the really quick responses. Here is a use case example:
int res = ftl::range(1, 4).map([](int a){
return ftl::range(a, 4).map([a](int b){
return std::make_tuple(a, b);
});
})
.flat_map([](std::tuple<int, int> x){ return std::get<0>(x) * std::get<1>(x); })
.sum();
assert(res, 25);
The ftl::range(begin, end) function returns a lazy iterator over the range [begin, end).

Answer 1

It's not that C++ disables the copy-assignment operator for lambda per-se, but that by default members in a lambda object are saved as const, and then the assignment operator can basically do nothing to assign to them, and so it is not generated. If you want lambdas to not hold members as const, you use the [...](...) mutable {...} syntax.
The other thing is that I'm not entirely sure what you get out of assigning lambdas. I mean, if you're going to re-use the lambda type (and functionality) and simply bind it to different variables, you're already working against the nice lambda capture syntax, and might as well have it be a normal function object. Assigning one type of lambda to another one is impossible. This means that you can not provide different lambda implementations when you hold the lambda itself by value.
If this is still what you're going for, I think dynamic allocation (e.g. using unique_ptr) is fair game.
And if you really want to avoid it, you could manually destruct and re-construct your lambda, as the following example illustrates:
#include <iostream>
template <class T>
struct LambdaContainer {
LambdaContainer(const T& lambda)
: lambda{lambda} {}
void resetLambda(const T& lambda) {
this->lambda.~T();
new (&this->lambda) T{lambda};
}
T lambda;
};
int main()
{
int i = 1;
auto l = [=]() {
std::cout << i;
};
using LT = decltype(l);
LambdaContainer<LT> lc{l};
lc.resetLambda(l);
}