Use of unary plus operator

Question

I've heard it is used as overloaded operator+ for example

class MyClass
{
    int x;
public:
    MyClass(int num):x(num){}
    MyClass operator+(const MyClass &rhs)
    {
        return rhs.x + x;
    }
};

int main()
{
    MyClass x(100);
    MyClass y(100);
    MyClass z = x + y;
}

Is this really the use of unary plus operator or is it really a binary + operator?

Where's the unary plus in the example? – Sz. Jul 25 '19 at 11:29 — Sz., Jul 25 '19 at 11:29

score 19 · Accepted Answer · answered Feb 25 '11 at 13:07

19

This is not overloading and using unary + .. You need to either make that a free function or make the member function take 0 arguments

class MyClass
{
    int x;
public:
    MyClass(int num):x(num){}
    MyClass operator+() const
    {
        return *this;
    }
};

int main() {
  MyClass x = 42;
  + x;
}

answered Feb 25 '11 at 13:07

Johannes Schaub - litb

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Could you return a 'reference' instead a value? – user963241 Mar 09 '11 at 16:45
4

@cpx yes but that would probably be a bad idea, because all types like `int` yield a value too, instead of a reference to the original variable. "Do it like the ints do" (Scott Meyers). – Johannes Schaub - litb Mar 09 '11 at 16:51

score 5 · Answer 2 · answered Feb 25 '11 at 13:07

5

That's a binary + operator. To overload the unary plus operator, you'd need something like this.

answered Feb 25 '11 at 13:07

Roger Lipscombe

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score 2 · Answer 3 · answered Feb 25 '11 at 13:09

2

When an operator-function is a member function, it has one more argument (this) than explicitly mentioned in the declaration. So in your case it's the binary +, the first argument is this, the second is the passed argument.

answered Feb 25 '11 at 13:09

Armen Tsirunyan

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CashCow · Answer 4 · 2011-02-25T13:14:29.800

1

This has been asked a few times before.

Your operator+ is binary. If defined as a member function needs to be const.

class MyClass
{
    int x;
public:
    MyClass(int num):x(num){}
    MyClass operator+(const MyClass &rhs) const
    {
        return rhs.x + x;
    }
};

It is quite often implemented in terms of operator +=

Note you have switched the commutativity but it will work as + for int is commutative.

A unary+ operator as a member would take no parameters and would not modify the object thus should be const. The implementation is up to your object. You could use it as "abs" for your int. (Using it as a no-op serves little purpose).

    MyClass MyClass::operator+() const
    {
        return ( x >= 0 ) ? MyClass( x ) : MyClass(-x);
    }

edited Feb 25 '11 at 13:14

answered Feb 25 '11 at 13:09

CashCow

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5

Using it as abs, on the other hand, would be just confusing. – UncleBens Feb 25 '11 at 13:17
1

Yeah but I see no point in overloading an operator to be a no-op albeit that it forces a copy. – CashCow Feb 27 '11 at 16:50
1

Then don't overload operator+. If you want absolute value, use a function that means absolute value. – David Stone Sep 01 '13 at 16:46

Use of unary plus operator

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