Quote:Original post by oconnellseanm

point a, b;
point &func(const point &a, const point &b)
{
point result = a + b;
return result;
}


I get an error for the line "point result = a + b"
Why can I not add 'a' and 'b'. I know they are constants, but is there a way to get around this?

Quote:Original post by Oluseyi

Quote:Original post by oconnellseanm
i have the following code

First off, defining the return type of your addition and subtraction overloads as a const non-reference means that they can not be chained (a + b + c - d would fail to compile).

The rest is a type mismatch.

Quote:Original post by Boku San
Also, compiler copy constructors don't ever work right in my experience...so I agree with Mr. Copy Constructor over there.

Quote:Original post by oconnellseanm
Here's the exact error from DevC++

189 C:\Documents and Settings\Sean\Desktop\main.cpp passing `const point' as `this' argument of `point point3::operator+(const point&)' discards qualifiers

Quote:Original post by Extrarius
Posting the exact error would be helpful. I'm going to have to disagree with nmi about the const: I've started labeling all return types const since they should be and I've had no such trouble.

foo a = b + c;

foo a = b + c * d;

Quote:Original post by snk_kid
Actually there is no gain from declaring the return value constant because first an expression such as:

foo a = b + c;

where the return type of the overloaded operator returns a constant foo, if the compiler doesn't optimize it will actually create an extra redundant copy because the type of the variable A in the assigment is not a constant type.

Quote:
So in the end there is no gain from declaring user-defined return type constant and may actually lead to redundant copying if not optimized by the compiler.

Quote:Original post by Boku San
compiler copy constructors don't ever work right in my experience...

Quote:Original post by Kaijin

Quote:Original post by snk_kid
Actually there is no gain from declaring the return value constant because first an expression such as:

foo a = b + c;

where the return type of the overloaded operator returns a constant foo, if the compiler doesn't optimize it will actually create an extra redundant copy because the type of the variable A in the assigment is not a constant type.

I'm not sure I understand. Are you saying that it wmight not perform the return value optimization if you try to assign a const to a non-const?

Quote:Original post by Kaijin

Quote:
So in the end there is no gain from declaring user-defined return type constant and may actually lead to redundant copying if not optimized by the compiler.

But isn't the point to declaring the return value as const to prevent the result being used as lvalue? i.e. (a + b) += c shouldn't be valid.