In my C++ book, I don't understand this particular example demonstrating operator overloading:
What I don't understand is the need to return *this. I know the function is specified as returning an object type 'ThreeD' but isn't that redundant? Would the following achieve the same thing:Code:ThreeD ThreeD :: operator=(ThreeD &op2) { x=op2.x; y=op2.y; z=op2.z; return *this; }
Code:void ThreeD :: operator=(ThreeD&op2) { x=op2.x; y=op2.y; z=op2.z; return; }
I AM WINNER!!!1!111oneoneomne
no it would not, because it is designed to assign. the = is the assignment operator, so it needs to return a ThreeD object, not void. if it does return void, then there would be no point to the operator.
basically, it will assign like this:
returning *this and having the function return a ThreeD object allows you to do this in main.Code:int main() { ThreeD object1, object2; object1 = object2; return 0; }
Ok thanks for your help. I tried it with void and it behaved as predicted. I think the reason why the book used *this was to allow compound assignments like a=b=c=d; to occur.
I AM WINNER!!!1!111oneoneomne
okay. thanks.
