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Operator overloading and namespaces

Asked 2010-10-08T14:11:07.940
13

Possible Duplicate:
Where should non-member operator overloads be placed?

While browsing on SO, I often find questions or answer that involves overloading/defining a std::ostream& operator<<(std::ostream& os, const Foo& foo) or a Foo operator+(const Foo& l, const Foo& r).

While I know how and when (not) to write these operators, I'm confused about the namespace thing.

If I have the following class:

namespace bar
{
  class Foo {};
}

In which namespace should I write the different operator definitions ?

// Should it be this

namespace bar
{
  std::ostream& operator<<(std::ostream& os, const Foo& foo);
}

// Or this ?

namespace std
{
  ostream& operator<<(ostream& os, const bar::Foo& foo);
}

// Or this ?

std::ostream& operator<<(std::ostream& os, const bar::Foo& foo);

The same question applies for the operator+. So, what is the good practice here and why ?

c++ namespaces operator-overloading operator-keyword
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2 Answers

15

The rule is that, when looking for a suitable function overload, both the current namespace and all namespaces of the argument type definitions are considered. This is called Argument Dependent Lookup (ADL).

So when you have this code:

  ::std::ostream& os = /* something */;
  const ::bar::Foo& foo = /* something */;
  os << foo;

The following namespaces are considered:

  • The current namespace
  • ::std, because os' type is defined there
  • ::bar, because foo's type is defined there

So all three possibilities you named, will work and thus are 'good enough' on first glance.

However....

You are not allowed to define new functions in ::std, so you can't put your overloaded operator in that namespace. (You are allowed to specialize templates in ::std, but that's not what we are doing here)

Secondly, the "current namespace" may change, so if you put your function definition in that namespace, it might not always be found.

So in the end, the best place to put the overloaded operator is in the same namespace as Foo:

namespace bar   
{   
  std::ostream& operator<<(std::ostream& os, const Foo& foo);   
}
answered 2010-10-08T14:23:17.857
0

The best choice is option 1. Why? Because when you use a unqualified function name (an an overloaded operator is a function), apart from normal name lookup, Argument-Dependent lookup is applied, that is (informally) all the namespaces where the arguments were declared are searched. E.g.

namespace N
{
   class X(){};
   void f(X){}
}
int main()
{
    N::X x;
    f(x); //works fine, no need to qualify f like N::f
}

The same is with operators.

On the other hand, in case of option 2 the operator still will be found because ostream is in std (same ADL rule). But it is not a good idea to add stuff to std namespace.

And the third option is bad, stylistically - why do it if the first option is sufficient?

So, definitely option 1.

HTH.

answered 2010-10-08T14:26:39.603

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