Why pass by const reference in overload operator+ with multiple parameters

问题

I am doing operator+ overloading with multiple parameters as below.

#include <iostream>
using namespace std;
class Integer{
    int value;
    public:
        Integer(int i) {value=i;};
        int getValue() { return value;};
        friend Integer operator+ (Integer & a, Integer & b){
            Integer I (a.value+b.value);
            return I;
        };
};

int main() {
    Integer a(1), b(2), c(3);
    Integer d = a+b+c;
    cout<<d.getValue()<<endl;
    return 0;
}

It can't be compile and return" no match for operator+". I read and understand the algorithm of multiple parameter that ((a+b)+c). Why it does not work? However, I found two ways to make it work:

friend Integer operator+ (const Integer & a,const Integer & b){
    Integer I (a.value+b.value);
    return I;
};

And

friend Integer & operator+ (Integer & a,Integer & b){
    Integer I (a.value+b.value);
    return I;
};

But I dont know why. Thank you

回答1:

Look at your operator+ signature:

friend Integer operator+ (Integer & a, Integer & b)
//                        ^^^^^^^^^    ^^^^^^^^^

a and b are lvalue references.

When you write

Integer d = a+b+c;

a+b produces an rvalue of type Integer, which does not bind to Integer&.

The version with const Integer & works as const lvalue references can be bound to both lvalues and rvalues.

回答2:

Executing a+b+c; will first invoke operator + with a and b which will produce a temporary Integer object, then invoke operator + with that temporary object and c. Calling operator + second time won't be possible because a temporary object can not be passed as reference to a non-const-qualified object.

Declaring operator + to accept non-const references does not make sense anyway because objects passes are not modified.

来源：https://stackoverflow.com/questions/47773497/why-pass-by-const-reference-in-overload-operator-with-multiple-parameters

标签

c++

operator-overloading

const

pass-by-reference