If the reference is returned, the caller will directly access the return value.
Usually the reference will point to the reference passed to the function, so calling the function is actually directly accessing one of its own variables.
take for example
const？ int & amp？ Fun? (int & amp？ First,? const？ int & amp？ b)
{
Answer? =? Answer? +? b；
Return? a；
}
int？ x？ =? 1,? y？ =? 2,? z；
z？ =? Fun (x,? y)；
//equivalent to? Fun (x,? y)； ? z？ =? x；
And returning a reference cannot return a temporary variable inside a function. Because the variable is destroyed at the end of the function. The return value will be meaningless. such as
const？ int & amp？ Fun? (int & amp？ First,? const？ int & amp？ b)
{
int？ c？ =? Answer? +? b；
Return? c；
}//The program will crash. Because the variable c is destroyed with the end of the program. Then the object with the same name returned in C is meaningless.