In C++, what does & mean after a function's return type?

Question

In a C++ function like this:

int& getNumber(); 

what does the & mean? Is it different from:

int getNumber(); 

Answer 1

It's different.

int g_test = 0;  int& getNumberReference() {      return g_test; }  int getNumberValue() {      return g_test; }  int main() {     int& n = getNumberReference();     int m = getNumberValue();     n = 10;     cout << g_test << endl; // prints 10     g_test = 0;     m = 10;     cout << g_test << endl; // prints 0     return 0; } 

the getNumberReference() returns a reference, under the hood it's like a pointer that points to an integer variable. Any change applyed to the reference applies to the returned variable.

The getNumberReference() is also a left-value, therefore it can be used like this:

getNumberReference() = 10; 

Answer 2

Yes, the int& version returns a reference to an int. The int version returns an int by value.

See the section on references in the C++ FAQ