Re: can the member function be compared?

"Alex Blekhman" <>
Fri, 26 Oct 2007 13:07:31 +0200
"Bill Gates" wrote:

But I found, the code like the following line can be
bool fRet = (void*)(&(&Foo::Func1)) <

So we can set a void pointer's value to be the reference
of the member function like below
void* ptr = ( void* )( &( &Foo::Func1 ) );
Now, I have a question about the pointer ptr. What dose
the pointer ptr point to?
In the standard, is this defined? (Sorry, I have no C++
ISO standard doc)

The value of `ptr' is undefined. Pointer to member is not
like regular pointer. Think about it as a struct. It can be
different in size. Moreover, pointer to member may not
contain an address at all. The only requirement for pointer
to member is to be able to call class' method via operators
`.*' and `->*'. Also, it should be comparable to NULL.

I mistakenly assumed in my previous post that pointers to
member of the _same class_ can be compared one with another
anyway, since a compiler could infer the real address of
method's body. However, Igor corrected me.

If you need to distinguish between pointers to member
function, then you could make arbitrary index for them. Like

struct X
    enum OP { OP_A, OP_B, ... };

    void opA(int param1, int param2);
    void opB(int param1, int param2);

    typedef std::map<OP, void (X::*)(int, int)> OPMap;

    OPMap m_opMap;

Then you could populate the map like this:

m_opMap[OP_A] = &X::opA;
m_opMap[OP_B] = &X::opB;


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