为什么转换构造函数和运算符重载+时用引用作为形参时c3=c1+1；报错而不用引用作为运算符重载形参时则编译通过

Question

#include
using namespace std;
class Complex
{
int r,i;
public:
Complex(){r=0;i=0;}
Complex(int t){r=t;i=0;}
Complex(int t,int m):r(t),i(m){}
friend Complex operator + (Complex t1,Complex t2);
friend ostream& operator << (ostream&,Complex&);
};
Complex operator + (Complex t1,Complex t2)
{
return Complex(t1.r+t2.r,t1.i+t2.i);
}
ostream& operator << (ostream &output,Complex &t)
{
if(t.i>0)
output<<t.r<<"+"<<t.i<<"i"<<endl;
else if(t.i==0)
output<<t.r<<endl;
else if(t.i<0)
output<<t.r<<t.i<<"i"<<endl;
return output;
}
int main()
{
Complex c1(1,2),c2(2,-2),c3;
c3=c1+c2;
cout<<c3;
c3=c1+1;
cout<<c3;
}
为什么把friend Complex operator + (Complex t1,Complex t2);改为friend Complex operator + (Complex& t1,Complex& t2);就会在c3=c1+1;这里出错？

Answer 1

1是常数不是变量，不能直接作为变量引用。Complex& t2是变量的引用，是告诉调用者，t2的值可能在函数中被改变。
程序可以改成
friend Complex operator + (const Complex& t1, const Complex& t2);
意思是函数中t1、t2都不会改变，1赋值改参数2,会创建一个新的Complex对象t2