Re: can the member function be compared?

"Alex Blekhman" <>
Thu, 25 Oct 2007 19:36:28 +0200
"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.


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