What does “int& foo()” mean in C++?

Question

While reading this explanation on lvalues and rvalues, these lines of code stuck out to me:

int& foo();
foo() = 42; // OK, foo() is an lvalue

Answer 1

In that context the & means a reference - so foo returns a reference to an int, rather than an int.

I'm not sure if you'd have worked with pointers yet, but it's a similar idea, you're not actually returning the value out of the function - instead you're passing the information needed to find the location in memory where that int is.

So to summarize you're not assigning a value to a function call - you're using a function to get a reference, and then assigning the value being referenced to a new value. It's easy to think everything happens at once, but in reality the computer does everything in a precise order.

If you're wondering - the reason you're getting a segfault is because you're returning a numeric literal '2' - so it's the exact error you'd get if you were to define a const int and then try to modify its value.

If you haven't learned about pointers and dynamic memory yet then I'd recommend that first as there's a few concepts that I think are hard to understand unless you're learning them all at once.