Why does operator+= only have one parameter?

[Halo Fan]

Here is an example I was given of a different type of code with the operator+= :

String& String:: operator+= (const String& s)
{
	int newlen = strlen(info_) + strlen(s.info_) + 1;

	char* temp = new char[newlen]; // Create new array for combined strings.
	strcpy_s(temp, newlen, info_); // Copy characters from this array into temp.
	strcat_s(temp, newlen, s.info_); // Append characters of s.info_ onto temp.

	delete [] info_; // Replace the old value for info_ by temp.
	info_ = temp;
	return *this;
}

Now I need to code my own operator+= method for a different kind of class but I am confused as to why it only gets one parameter. doing operator+ made sense to me because I just had to find a way to get x + y to work. I guess I am not sure what I am even trying to do with operator+= though. Here is an outline of the function I am working on:

	LongInt& LongInt::operator+=(const LongInt& rhs)
	{
		
	}

Thank you for your time in reading this.

I solved it and here is what I came up with:

	LongInt& LongInt::operator+=(const LongInt& rhs)
	{
		LongInt temp;
		temp = *this;
		temp = temp + rhs;
		*this = temp;

		return *this;
	}

This link helped me a lot in understanding what it was I was trying to do :P
http://courses.cms.caltech.edu/cs11/material/cpp/donnie/cpp-ops.html

[Disch]

I'm not sure, but it sounds like you answered your own question. Is that true? Or did you still need clarification?


Anyway, there's no need to create a temporary variable. That just makes your code a little longer and more confusing. Just modify 'this' directly:

	LongInt& LongInt::operator+=(const LongInt& rhs)
	{
		*this = *this + rhs;
		return *this;
	}