Return by Reference

[anhnha]

Hi,
Here is the problem:
From the page: http://www.programiz.com/cpp-programming/return-reference

#include <iostream>
using namespace std;
int n;
int& test();

int main() {
    test() = 5;
    cout<<n;
    return 0;
}

int& test() {
    return n;
}

Explanation

In program above, the return type of function test() is int&. Hence this function returns by reference. The return statement is return n; but unlike return by value. This statement doesn't return value of n, instead it returns variable n itself.

Then the variable n is assigned to the left side of code test() = 5; and value of n is displayed.

I don't quite understand the bold sentence.
Shouldn't value of n and variable n be the same?

[poteto]

If it was just the variable, doing test() = 5; would be like 0=5;(I don't believe it would make an error, but probably a warning), and with the reference"&", its like n = 5;.

[mutexe]

it means it's not returning the value of n, it's returning where n is sitting in memory so it can be assigned a value (5 in your case).

your main() is the equivalent of this:

int n = 5;
cout << n;

[anhnha]

Thanks for replies!
With the declaration, is n a variable or value?

int n;

As I understanding, n represents both a variable and its value.

[JLBorges]

int n = 5 ; // 'n' is a variable; it denotes an object of type int

int& r = n ; // 'r' is a variable; it denotes a reference to int (an alias for an object of type int)
             // in this case, 'r' is an alias for the object of type int denoted by the variable 'n'  

When evaluated as an expression, both (n) and (r) yield a value (an lvalue) of type int.

[anhnha]

Sorry, I am still confused about returning a reference.
In the case above, the test() function will return a variable &n and this is an alias for n?
Then the alias is assigned with the value 5, meaning that &n = 5.
So, n = 5.

P.S:
I think here is what I am confused about. Let's consider the assignment below:
a = 3;

What it does here is assign rvalue of (3) to lvalue of (a). Is that right?

[JLBorges]

With this function:

int foo() // return type is not a reference
{
    int temp = 5 ;
    return temp ;
}

The expression foo() is a prvalue (pure rvalue) of type int.
The prvalue, in this case is a value not associated with any object.
In other cases, a prvalue (for instance a prvalue of type std::string) may identify a temporary object.


With this function:

int& bar() // return type is (lvalue) reference to int
{
    static int i = 5 ; // static storage duration
                       // life-time of the object does not end when the function returns
    return i ;
}

The expression bar() is an lvalue of type int
The lvalue identifies a non-temporary object of type int

[anhnha]

Yes, that makes more sense now. I have to read up about lvalue, rvalue and reference type returning functions.

[JLBorges]

Value categories are explained here: http://en.cppreference.com/w/cpp/language/value_category