需要帮助空转，不知道输出如何

Question

I'm having difficulty dry running this code. I can understand clearly whats going on till:

Line 8

Line 8

Line 9

Line 7

Line 8

Line 1

but after this things get muddy. Can someone explain just a little bit please? How are there three Line 7s after the above? Shouldn't there be two?

Sorry its a bit long but I'm at a loss to see how I'm not getting it right.

Code:

#include <iostream>
using namespace std;

class Square {
    friend Square operator -(Square sq1, Square sq2);
    friend ostream& operator<<(ostream& out, Square & s);
private:
int side;
public:
Square(int);
        Square(Square &);
        Square operator+(Square);
        Square operator/(Square);
        Square operator ++(int);
        Square operator--();
        int getSide();
int getArea();
};

Square::Square(int _side){
   cout<<"Line 8"<<endl;                   
side = _side;
}
Square::Square(Square & s){
   cout<<"Line 7"<<endl;
side = s.side;
}

Square Square :: operator +(Square sq){
   Square squareTemp(0);
squareTemp.side = side + sq.side/sq.side - 1;
cout<<"Line 1"<<endl;
return squareTemp-sq-sq;
}
Square Square :: operator /(Square sq){
    Square squareTemp(0);
squareTemp.side = side - 1;
cout<<"Line 2"<<endl;
return --squareTemp;
}

Square Square :: operator ++(int i){
    Square squareTemp(0);
squareTemp.side = side + 2;
cout<<"Line 3"<<endl;
return squareTemp;
}
Square Square :: operator --(){
    Square squareTemp(0);
squareTemp.side = (squareTemp.side++) +side + 1;
cout<<"Line 4"<<endl;
return squareTemp;
}

int Square :: getSide(){
    cout<<"***If this is the last line of output then it is Wrong Result"<<endl;
return side;
}

int Square :: getArea(){
    cout<<"Line 5"<<endl;
side++;
return side * side;
    cout<<"Line 21"<<endl;
}

Square operator-(Square sq1, Square sq2){
    Square squareTemp(0);
squareTemp.side = sq1.side + sq2.side;
cout<<"Line 6"<<endl;
return squareTemp;
}

ostream&operator<<(ostream& out, Square & s){
    out<<"s.side"<<endl;
return out;
}


int main(){
Square s1(3);
Square s2(4);
cout<<"Line 9"<<endl;

Square s = s1 + s2;
cout<<"Side of s:"<<s.getSide() <<endl;
cout<<"Area of s:"<<s.getArea() <<endl;
cout<<s1<<endl;
cout<<s2<<endl;
cout<<s.getSide()<<endl;
system("PAUSE");
}

Output:

Line 8

Line 8

Line 9

Line 7

Line 8

Line 1

Line 7

Line 7

Line 7

Line 8

Line 6

Line 7

Line 8

Line 6

Line 7

***If this is the last line of output then it is Wrong Result

Side of s:11

Line 5

Area of s:144

s.side

s.side

***If this is the last line of output then it is Wrong Result

12

Answer 1

The line you're getting confused on is bit from operator+ .

cout<<"Line 1"<<endl;
return squareTemp - sq - sq;

The sequence of steps is

1. evaluate squareTemp - sq , which matches Square::operator-(Square sq1, Square sq2)
2. copy-construct argument sq1 from squareTemp - output "Line 7"
3. copy-construct argument sq2 from sq - output "Line 7"
4. Output "Line 7" (I will explain this in a moment)
5. int-construct squareTemp in operator- - output "Line 8"
6. operator- outputs "Line 6"
7. Copy-construct the return value of operator- from `squareTemp - output "Line 7"
8. Now we have the second part of (squareTemp-sq) - sq which is Square::operator-(Square sq1, Square sq2)
   1. The return value is used as sq1 - in theory another copy could be done, but the compiler elides this.
   2. The object created in step 4 is sq2 here
9. (Repeat steps 5,6,7 to output "Line 8" "Line 6" "Line 7"

The reason that the step 4 is allowed to be created there even though it isn't used until step 8, is because there is no sequence point in the expression:

Square::operator-( Square::operator-(squareTemp, sq), sq )

The compiler must evaluate squareTemp and the first sq before calling the inner operator- . However, it is free to evaluate the second sq at any point before the call to the second operator- .