Re: can the member function be compared?

"Igor Tandetnik" <>
Thu, 25 Oct 2007 13:50:07 -0400
Alex Blekhman <> wrote:

"Bill Gates" wrote:

for example, there is a class as below

struct Foo
void Func1()
void Func2()

As known, &Foo::Func1 != &Foo::Func2 is right, but
&Foo::Func1 < &Foo::Func2
is illegal. So std::map< void ( Foo::* )() > can not be
defined. Are there
some other ways to get the greater and less relationship
between two member

The "&Foo::Func1 < &Foo::Func2" expression is perfectly
legal if both `Func1' and `Func2' are within the same access
level block. I.e., if `Func1' and `Func2' are not separated
by "public", "protected" or "private" labels, then their
comparison is legal and valid.

Here's the quote from the Standard:

5.9/2 Relational operators

If two pointers point to nonstatic data members of the
same object, or to subobjects or array elements of such
members, recursively, the pointer to the later declared
member compares greater provided the two members are not
separated by an `access-specifier' label (11.1) and provided
their class is not a union.
If two pointers point to nonstatic data members of the
same object separated by an `access-specifier' label (11.1)
the result is unspecified.

Note that this paragraph talks about regular pointers to data members.
It's not applicable to pointers-to-members. 5.9/1 is applicable though:

5.9/1 The operands shall have arithmetic, enumeration or pointer type.

Note further that, according to 3.9.2/1, pointer types and
pointers-to-member types are different kinds of types.

If you are still not convinced, look at the stadnard definition for
equality (==) operator:

5.10/1 The == (equal to) and the != (not equal to) operators have the
same semantic restrictions, conversions, and result type as the
relational operators...
5.10/2 In addition, pointers to members can be compared...

Bottom line - pointers-to-members cannot be compared for inequality.
With best wishes,
    Igor Tandetnik

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