What does "int& foo()" mean in C++? - 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
I tried it in g++, but the compiler says "undefined reference to foo()". If I add
int foo()
{
return 2;
}
int main()
{
int& foo();
foo() = 42;
}
It compiles fine, but running it gives a segmentation fault. Just the line
int& foo();
by itself both compiles and runs without any problems.
What does this code mean? How can you assign a value to a function call, and why is it not an rvalue?

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

All the other answers declare a static inside the function. I think that might confuse you, so take a look at this:
int& highest(int & i, int & j)
{
if (i > j)
{
return i;
}
return j;
}
int main()
{
int a{ 3};
int b{ 4 };
highest(a, b) = 11;
return 0;
}
Because highest() returns a reference, you can assign a value to it. When this runs, b will be changed to 11. If you changed the initialization so that a was, say, 8, then a would be changed to 11. This is some code that might actually serve a purpose, unlike the other examples.

Answer 3

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 4

int &foo(); declares a function called foo() with return type int&. If you call this function without providing a body then you are likely to get an undefined reference error.
In your second attempt you provided a function int foo(). This has a different return type to the function declared by int& foo();. So you have two declarations of the same foo that don't match, which violates the One Definition Rule causing undefined behaviour (no diagnostic required).
For something that works, take out the local function declaration. They can lead to silent undefined behaviour as you have seen. Instead, only use function declarations outside of any function. Your program could look like:
int &foo()
{
static int i = 2;
return i;
}
int main()
{
++foo();
std::cout << foo() << '\n';
}

Answer 5

int& foo(); is a function returning a reference to int. Your provided function returns int without reference.
You may do
int& foo()
{
static int i = 42;
return i;
}
int main()
{
int& foo();
foo() = 42;
}

Answer 6

int & foo(); means that foo() returns a reference to a variable.
Consider this code:
#include <iostream>
int k = 0;
int &foo()
{
return k;
}
int main(int argc,char **argv)
{
k = 4;
foo() = 5;
std::cout << "k=" << k << "\n";
return 0;
}
This code prints:
$ ./a.out
k=5
Because foo() returns a reference to the global variable k.
In your revised code, you are casting the returned value to a reference, which is then invalid.

Answer 7

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.

Answer 8

The example code at the linked page is just a dummy function declaration. It does not compile, but if you had some function defined, it would work generally. The example meant "If you had a function with this signature, you could use it like that".
In your example, foo is clearly returning an lvalue based on the signature, but you return an rvalue that is converted to an lvalue. This clearly is determined to fail. You could do:
int& foo()
{
static int x;
return x;
}
and would succeed by changing the value of x, when saying:
foo() = 10;

Answer 9

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! :)