Re: assignment operator

=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?= <>
Sun, 23 Sep 2007 12:30:46 GMT
On 2007-09-23 13:32, Michal Nazarewicz wrote:

On 2007-09-23 12:28,, India wrote:

For a class Test, we write the assignment operator as

Test & Test::operator=(const Test & rhs);

instead of

const Test & Test::operator=(const Test & rhs);

that is, the return type of operator=( ) is just a reference, not a
reference to const object.

is it because to allow for expressions such as

(x = y) + z;
where x, y, z are of type Test ?

What would happen if we allowed "const Test &" as the return type of
Test::operator=( ) ?

Erik Wikstr??m <> writes:

I would say it is because there is no reason to return a const reference
(since the object referred to is not const) and it does allow us to do
some things that a const reference would not. Your example above "(x =
y) + z;", however, should work even with a const reference,

How come? There is no "operator=(const Test&)const", so how could it

Because the plus operator should take const references as arguments and
be declared as const.

class Foo
  const Foo& operator=(const Foo& rhs) { return *this; }
  Foo operator+(const Foo& rhs) const { return Foo(); }

int main()
  Foo a, b, c;

  Foo d = (a = b) + c;

  Foo e = a + b;

following would not "(x = y) += z;", but on the other hand you should
probably write such code in the first place. Notice also that the other
unary operators, such as +=, *=, etc., return a non-const reference.

Those are not unary operators and they return the type you specify which
should be const reference.

Sorry, yes they are not unary operators, I meant assignments operators.
And of course they return whatever type you want them to return, but
they *should* return a non-const reference to themselves (or a type
which is convertible to that).

Programmers don't expect expressions such as "(a = b) = c" or "(a += b)
= c" to work and thus they should not work. Operator overloading is not
for creating new syntax.

I expect them to work, the most important thing when overloading
operators is to keep the semantics sane, which usually means that it
should be compatible with the semantics of built-in types. Of course it
is really bad style to actually use such code, but it should work.

Erik Wikstr??m

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