Re: why not implicit operator !=()?

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <>
Thu, 15 Mar 2012 13:42:09 -0700 (PDT)
On 2012-03-15 08:07, Michael wrote:

This has probably been asked before but...

Is there a particular reason why when you define an operator ==()
for a type:

struct SomeType
   bool operator== (const SomeType& rhs) const { return ...; }

... that the compiler doesn't also go ahead and implicitly define
operator !=() for you?

Please no! First, this would silently change the semantics of existing
programs. Implicit member declarations are hard to explain, and during
the C++11 standardization one of the previous implicit member
declarations has now been deprecated: If a class type has a
user-declared copy-assignment operator [copy constructor] or destructor,
the implicit copy constructor [copy-assignment operator] declaration is
now deprecated.

This omission has become a pet peeve of mine lately...

I understand that and I would be much in favour for extending the
concept of defaulted member functions. I would very much like your
example if you had suggested

struct SomeType
   bool operator==(const SomeType& rhs) const { return ...; }
   bool operator!=(const SomeType& rhs) const = default;

for example. Extending the possibility of defaulted functions was an
early idea during the C++11 standardization process, see e.g.

and the ==/!= example has already been mentioned in

But they were simply too late for C++11. IMO extending defaulted
functions to





would be a great idea. But it is important to get the semantics right: I
expect that if I write

struct SomeType
   bool operator==(const SomeType& rhs) const { return ...; }
   bool operator!=(const SomeType& rhs) const = default;
} s1, s2;

the semantics of s1 != s2 is exactly that of !(s1 == s2), but what if
you write

struct SomeType
   bool operator!=(const SomeType& rhs) const = default;
} s1, s2;

? Does this define member-wise inequality? Another more severe problem
to solve would be a mixture of free and member functions. Consider:

struct SomeType
   int a, b;
   bool operator!=(const SomeType& rhs) const = default;
   // return (a == rhs.a) && (b == rhs.b) ??
} s1, s2;

bool operator==(const SomeType& lhs, const SomeType& rhs) { return lhs.a
== rhs.a; }

// User says: SomeType::b is no member contributing to ==

Unfortunately these examples reveal some serious problems to solve.

HTH & Greetings from Bremen,

Daniel Kr?gler

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