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Question

Our Professor has given us a practice test with the answers to help us for our upcoming test. I don't understand how the answer for this code is 135 .

I understand what a class is but I am struggling how operators work. For v{6} I understand that for the object v the int v equals 13 . I thought that in the int main that the first -v in (-v - v).print would be evaluated first in the operator that return 2 * v and that it would return 26 . I then thought they would then be put in the last operator

V operator-(int lop, const V& rop)
{
    return rop - lop;
}

but I don't think that is the case. I have put this code into Visual Studio to mess around with it but I don't understand what is going on.

#include <iostream>
using namespace std;

class V
{
    int v;

public:
    V(int a = 3, int b = 7) : v{a + b}
    {
    }

    void print()
    {
        cout << v;
    }

    V operator-(const V& rop) const
    {
        return (3 * v) + (2 * rop.v) + 3;
    }

    V operator-()
    {
        return 2 * v;
    }
};

V operator-(int lop, const V& rop)
{
    return rop - lop;
}

int main() 
{
    V v{6};
    (-v - v).print();

    return 0;  
}

Answer 1

Wow, this is confusing code, with re-used variable names, strange unconventional operations and such. The code is particularly hard to follow because V can be implicitly constructed from an integer, with 7 always added to that integer; even with a debugger, this took me a few moments to grok. Please never write code like this, not even for fun!

The result of -v in main is not an int . It is a V object implicitly constructed from the expression 2 * v (=26), resulting in a member integer with value 33 (26+7).

It's as if you wrote:

V operator-()
{
    return V(2 * v);
}

or, due to the default argument:

V operator-()
{
    return V(2 * v, 7);
}

Then you take this new returned object and feed it to member operator-(const V&) ; same story applies. It produces the expression 3*33 + 2*13 + 3 , which is 128; again this is used to construct a new V (because that's the return type!), so add 7 to get 135.

operator-(int lop, const V& rop) doesn't come into it because you never performed a subtraction between an int and a V .

Answer 2

The result of -v is not an int , but a V .
Thus, the member-overloaded subtraction will be used.

This is the equivalent code, without the syntactic sugar of operators, but with explicit conversion of return values:

class V
{
public:
    V(int a = 3, int b = 7) : v{a + b}
    {
    }

    void print()
    {
        cout << v;
    }

    V subtract(const V& rop) const
    {
        return V((3 * v) + (2 * rop.v) + 3);
    }

    V negate()
    {
        return V(2 * v);
    }
private:
    int v;
};


int main() 
{
    V v{6};
    (v.negate().subtract(v)).print();

    return 0;  
}