CODEWITHSUNDEEP
c++operator-overloading
Raulp
Raulp

Reputation: 8136

Overloaded Addition assignment operator in C++ for two /more than two objects?

I have overloaded the + operator like this 

class sample
{
private : 
  int x;
public :
  sample(int x1 =0)
  {
    x = x1;
  }

  sample operator+(sample s);
};

sample sample::operator+(sample s)
{
  x = x + s.x;
  return *this;
}

int  main()
{
  sample s1(10);
  sample s2;
  s2 = s2 + s1;
  return 0;    
}

Is this correct? My question is If I want to add two different sample objects how will I overloaded the opeartor; e.g for s = s1 + s2;

I feel like doing s = s + s1 + s2 with the existing implementation.

Upvotes: 19

Views: 58383

Answers (1)

dinox0r
dinox0r

Reputation: 16039

Using friend operator overload should do the trick for you and is a common way to define binary operators, just add:

friend sample operator+(const sample& a, const sample& b); //in class

sample operator+(const sample& a, const sample& b) { //outside the class
    return sample(a.x + b.x);
}

If you want it to remain a member, (which has downsides in some rare scenarios, and no upsides), you have to make the operator a const function:

sample operator+(sample s) const; //in class

sample sample::operator+(const sample& b) const { //outside the class
    return sample(this->x + b.x);
}

Either of these will allow operator chaining. The reason your previous s = s + s1 + s2 was failing, is that the s + s1 would execute and return a temporary sample object. Then, it would attempt to add s2 to that sample. However, temporaries can only be const references[1], and as such, can only use const member functions. Since your operator+ member function is not a const function, you cannot use that function on the const temporary. Note that to make it const, I had to rewrite it, since your version modifies the object on the left side of the +.

[1]with exceptions aren't particularly relevant here, namely rvalues

Upvotes: 32

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