Re: why not implicit operator !=()?

On 2012-03-15 08:07, Michael wrote:


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.


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.

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2008/n2584.html

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

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/n2210.html

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

==
!=
<

<=


swap

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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