c++ - What happens if I capture a local variable by reference, and it goes out of scope?

Question

Suppose I use a lambda as a callback function, and when creating the lambda, I capture a local function variable by reference. Now suppose that the lambda object does not get executed until after that local function variable goes out of scope. What happens?

I realize that it would be pretty stupid for someone to do so if there's a chance of it happening, but I am almost positive that someone would end up doing it.

Answer 1

Yes, that would be following a dangling reference. It sounds like you're worried about interface design: "I am almost positive that someone would end up doing it." Please don't reject lambdas and std::function on this basis, as they are no more dangerous than any other alternative. Lambdas are just a simpler way to define local functors. std::function is the best interface to persistent, polymorphic functors, lambda or not.

The scope issue is why it's easier to capture by value. The user won't get a reference unless they write &. Of course, the danger is that someone would get in the habit of starting all their lambda functions with [&], since references are "faster." Hopefully any such person would learn their lesson soon enough… although some pointer-happy folks are just incorrigible.