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

The explanation is assuming that there is some reasonable implementation for foo which returns an lvalue reference to a valid int.

Such an implementation might be:

int a = 2; //global variable, lives until program termination

int& foo() {
    return a;
} 

Now, since foo returns an lvalue reference, we can assign something to the return value, like so:

foo() = 42;

This will update the global a with the value 42, which we can check by accessing the variable directly or calling foo again:

int main() {
    foo() = 42;
    std::cout << a;     //prints 42
    std::cout << foo(); //also prints 42
}

Answer 2

int& foo();

Declares a function named foo that returns a reference to an int. What that examples fails to do is give you a definition of that function that you could compile. If we use

int & foo()
{
    static int bar = 0;
    return bar;
}

Now we have a function that returns a reference to bar. since bar is static it will live on after the call to the function so returning a reference to it is safe. Now if we do

foo() = 42;

What happens is we assign 42 to bar since we assign to the reference and the reference is just an alias for bar. If we call the function again like

std::cout << foo();

It would print 42 since we set bar to that above.

Answer 3

The function you have, foo(), is a function that returns a reference to an integer.

So let's say originally foo returned 5, and later on, in your main function, you say foo() = 10;, then prints out foo, it will print 10 instead of 5.

I hope that makes sense :)

I'm new to programming as well. It's interesting to see questions like this that makes you think! :)