From: horst DOT kraemer AT t-online DOT de (Horst Kraemer)
Newsgroups: comp.os.msdos.djgpp
Subject: Re: Possible Compiler Bug
Date: Sat, 26 Feb 2000 11:33:09 GMT
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References: <38B57379 DOT 817FFBE4 AT netcom DOT ca>
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On Thu, 24 Feb 2000 13:07:53 -0500, MM <mm AT netcom DOT ca> wrote:

> #include <iostream.h>
> 
> class t
> {
>    private:
>       int a;
>    public:
>       t (void) : a (10) { };
> 
>       void selectfunc (int w);
>       void multifunc (void);
>       void (t::*funcptr)(void);
> 
>       void func1 (void);
>       void func2 (void);
>       void func3 (void);
> };
> 
> void t::func1 (void) { cout << "Func1 " << a << endl; }
> void t::func2 (void) { cout << "Func2 " << a << endl; }
> void t::func3 (void) { cout << "Func3 " << a << endl; }
> 
> void t::selectfunc (int w)
> {
>    switch (w)
>    {
>       case 1: funcptr = &func1; break;
>       case 2: funcptr = &func2; break;
>       case 3: funcptr = &func3; break;
>    }
> }
> 
> void t::multifunc (void)
> {
>    this->funcptr ();
> }
> 
> int main (void)
> {
> 
>    return (0);
> }


1) It is certainly a bug if a compiler crahes upon an unexpected
situation and illegal code.


2) As to your question if you are doing something wrong:

The statement

	this->funcptr();

is ill-formed and meaningless. What you want is

	
void t::multifunc (void)
{
   (this->*funcptr)();
}


If you want to call a member function through an object and a pointer
to a member function you have to apply the operator .* (for objects)
or ->* for pointers to objects and you have to put the expression

	(obj.*membfuncptr) or (objptr->*membfuncptr)

into brackets because the operators .* and ->* have a lower
precendence than the function call operator () - while the "normal"
operators . and -> have the same precedence as (). Thus

	this->func()

would be interpreted automatically as

	(this->func)()

- which is what you mean - but

	this->*funcptr()

would be interpreted as

	this->*(funcptr())

which is (usually) not what you mean. Note that .* and ->* are not
"combinations" of the operators (. -> *). They are independent
operators of their own and don't inherit properties from their lexical
components (. -> *).

And please note that ->* or .* do _not_ automatically introduce the
scope of the left operand like the operators (. ->). I.e. in this
context

struct X
{
   void (X::*mfp)();
};

void f()
{
  X* px = new X;
  
  (px->*mfp)();  // Error
}

would be an error because there is no mfp in scope. The correct call
in order to specify the pointer px->mfp would be

  (px->*px->mfp)();


Regards
Horst