Why Define Operator + or += Outside a Class, and How to Do It Properly

Why define operator + or += outside a class, and how to do it properly?

The first form of the operators is what you would define inside class Type.

The second form of the operators is what you would define as free-standing functions in the same namespace as class Type.

It's a very good idea to define free-standing functions because then the operands to those can take part in implicit conversions.

Example

Assume this class:

class Type {
    public:
    Type(int foo) { }

    // Added the const qualifier as an update: see end of answer
    Type operator + (const Type& type) const { return *this; }
};

You could then write:

Type a = Type(1) + Type(2); // OK
Type b = Type(1) + 2; // Also OK: conversion of int(2) to Type

But you could NOT write:

Type c = 1 + Type(2); // DOES NOT COMPILE

Having operator+ as a free function allows the last case as well.

What the second form of the operator does wrong though is that it performs the addition by directly tweaking the private members of its operands (I 'm assuming that, otherwise it would not need to be a friend). It should not be doing that: instead, the operators should also be defined inside the class and the free-standing functions should call them.

To see how that would turn out, let's ask for the services of a guru: http://www.gotw.ca/gotw/004.htm. Scroll at the very end to see how to implement the free-standing functions.

Update:

As James McNellis calls out in his comment, the two forms given also have another difference: the left-hand-side is not const-qualified in the first version. Since the operands of operator+ should really not be modified as part of the addition, it's a very very good idea to const-qualify them all the time. The class Type in my example now does this, where initially it did not.

Conclusion

The best way to deal with operators + and += is:

1. Define operator+= as T& T::operator+=(const T&); inside your class. This is where the addition would be implemented.
2. Define operator+ as T T::operator+(const T&) const; inside your class. This operator would be implemented in terms of the previous one.
3. Provide a free function T operator+(const T&, const T&); outside the class, but inside the same namespace. This function would call the member operator+ to do the work.

You can omit step 2 and have the free function call T::operator+= directly, but as a matter of personal preference I 'd want to keep all of the addition logic inside the class.

Operator overloading outside class

The basic question is "Do you want conversions to be performed on the left-hand side parameter of an operator?". If yes, use a free function. If no, use a class member.

For example, for operator+() for strings, we want conversions to be performed so we can say things like:

string a = "bar";
string b = "foo" + a;

where a conversion is performed to turn the char * "foo" into an std::string. So, we make operator+() for strings into a free function.

operator overloading outside of a class!

I suspect you have the definition in a different namespace than the declaration. ADL is finding the declaration (since it's in the same namespace as the class), and then you get an unresolved external error during link.

e.g.

-foo.h-

namespace aspace
{
  class A
  {
  public:
      A(float x);
      float x;
  };
  A operator +(const A&, const A&);
}

-foo.cpp-

#include "foo.h"
using namespace aspace;

A::A(float x)
{
    this->x = x;
}

A operator +(const A& lh, const A& rh)
{
    return A(lh.x + rh.x);
}

Will give the error you describe. The solution is to put the operator+ definition in the correct namespace:

namespace aspace
{
  A operator +(const A& lh, const A& rh)
  {
      return A(lh.x + rh.x);
  }
}

Overload + Operator for Class Data

1. To answer your first 2 questions

   It is recommended (but not enforced) to use the friend approach for binary operators because it allows to have objects that are convertible to your class on the left hand side of the operator. Consider this:

   class Foo
   {
       // private members
   public:
       Foo(int) {/* ctor implementation */} // implicit convertible to int
       Foo(const Foo&) { /* copy ctor implementation */ }
       friend Foo operator+(const Foo& lhs, const Foo& rhs){/* implementation */}
   };
   

   Now you can do:

   Foo foo1{1};
   Foo foo2 = 1 + foo1; // 1 is implicitly converted to Foo here
   

   If operator+ would have been a member function, the call above would fail (left hand side, even if convertible to Foo, will not be automatically converted by the compiler), and you'd be only able to use

   Foo foo2 = foo1 + 1;
   

   So, in conclusion, using friend for binary operators make them more "symmetric".

2. Last question:

   The assembler would not generate the same code, as the friend/member operator+ are doing slightly different things.

See also a very good guide about overloading operators here.

Can comparison operator be defaulted outside of class definition in C++20?

P2085R0 removed the requirement on the defaulted comparison operator to be defaulted on the first declaration. Clang currently doesn't support this proposal:

See also https://reviews.llvm.org/D103929

C# Overloading operator== outside the class

No, it's not possible to override an operator from a class that is not involved in the operation.

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You can make a class that implements IEualityComparer<SomeClass>, which can be used instead of the standard comparison in some cases, for example in a dictionary:

var x = new Dictionary<SomeClass, string>(new SomeClassEqualityComparer());

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If you just want to use the comparison in your own class, you could make it a regular static method instead of overriding an operator:

public static bool SomeClassEqual(SomeClass a, SomeClass b) { 
  if (a.i == b.i) {
    return true;
  }
  // compare some other members as well
  return false; 
}

Usage:

if (SomeClassEqual(a, b))

C++ Defining the << operator of an inner class

Since the friend operator is first declared inside the class, it's only available by argument-dependent lookup. However, neither of its parameter types are in namespace A, so it won't be found. C is an alias for std::map, so is considered to be in namespace std for the purposes of ADL.

There are various ways you could fix it, none of which are perfect:

• Declare the function in namespace A before the class definition; then it becomes available by normal lookup, not just ADL. However, this breaks the encapsulation somewhat, and might cause problems if anything else tries to overload operator<< for std::map.
• Replace the operator overload with a named static (not friend) function, and call it by name.
• Declare C as an inner class, rather than an alias for std::map. This enables ADL without breaking encapsulation, but is a bit awkward if you want it to behave just like std::map.

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