What does =+ mean in c++

Question

I've been studying operator overloading and i cant understand this, whenever i use s1=+s2 answer is s1=1 s2=1 and when i use s2=+s1 i get s1=2 s2=2 please explain

#include<iostream>
using namespace std;
class score
{   private:
    int val;
    public:
    score()
    {   val=0;  }
    score operator+()
    {   score temp;
        val=val+1;
        temp.val=val;
        return temp;    }
    void show()
    {   cout<<val<<endl;    }
};
main()
{
    score s1,s2;
    s1.show();
    s2.show();
    +s1;
    s1=+s2;
    s1.show();
    s2.show();
}

@sameeralakshitha That is incorrect, you are thinking of `s1 += s2;` This is not the same thing. It's equivalent to `s1 = (+s2);`, which is `s1 = s2;` in most cases. — François Andrieux, May 16 '17 at 19:24
@sameeralakshitha `s1=+s2;` is not the same as `s1 = s1 + s2;`. `s1=+s2; => s1 = +s2;` and `s1+=s2 => s1 = s1 + s2;` — NathanOliver, May 16 '17 at 19:24
Off topic: That + operator is violates convention. + should not modify the objects it's invoked on just like 3 = 2+1 doesn't modify the value of 2 or 1. More here: [Operator overloading](http://stackoverflow.com/questions/4421706/operator-overloading) — user4581301, May 16 '17 at 19:25

πάντα ῥεῖ · Accepted Answer · 2017-05-16T19:39:36.047

0

 s2=+s1

Means merely nothing more than applying the unary operator+() to the rvalue provided with the operator=()'s parameter.

That boils down to the same statement as

s2 = (+s1);

There's no combined operator like operator+=() for that.

That would have no effect at all, unless these operator functions are overloaded for specific (non primitive) types.

edited May 16 '17 at 19:39

answered May 16 '17 at 19:26

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I believe `operator+` is called first. From my understanding of your answer, it seems to imply the opposite. – François Andrieux May 16 '17 at 19:29
@FrançoisAndrieux The opposite of what? – πάντα ῥεῖ May 16 '17 at 19:29
Your answer says that `operator+` is called on the result of `operator=`, but it's the `operator=` that should be acting with the result of `operator+`. `s2.operator=(`s1.operator+()`);` – François Andrieux May 16 '17 at 19:31
The opposite of the truth; Francois is right, but I think it's just a language barrier with the way you phrased "provided with the operator=" – Lightness Races in Orbit May 16 '17 at 19:31
@Boundary I tried to clarify that now. – πάντα ῥεῖ May 16 '17 at 19:31
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It's a bit better. Prob good enough now (particularly since this is a dupe anyway) – Lightness Races in Orbit May 16 '17 at 19:33

What does =+ mean in c++

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