Re: [c++] Another question about demangler output

[Gabriel Dos Reis <gdr at integrable-solutions dot net>]

Daniel Jacobowitz <drow@mvista.com> writes:

| On Sun, Dec 07, 2003 at 05:24:46AM +0100, Gabriel Dos Reis wrote:
| > Ian Lance Taylor <ian@wasabisystems.com> writes:
| > 
| > | Gabriel Dos Reis <gdr@integrable-solutions.net> writes:
| > | 
| > | > Ian Lance Taylor <ian@wasabisystems.com> writes:
| > | > 
| > | > [...]
| > | > 
| > | > | This gives _ZN2BBcvPFivEEv which (currently) demangles as
| > | > |     BB::operator int (*)()()
| > | > | 
| > | > | But using a gcc extension, I can do this without a typedef:
| > | > | 
| > | > | class BB { operator typeof (int(*)())(); };
| > | > | BB::operator typeof (int(*)())() { return 0; }
| > | > 
| > | > If you do that, then you might end up accpeting two different
| > | > declarations as same where the token-oriented scheme (ODR) would have
| > | > kept them separate.  That is, you would not be able to differentiate
| > | > 
| > | > tu1.C:
| > | > 
| > | >   struct B { operator typeof(int(*)())(); };
| > | > 
| > | > from
| > | > 
| > | > tu2.C:
| > | > 
| > | >   struct B { typedef int (*foo)(); operator foo(); };
| > | > 
| > | > ODR says they are different.
| > | 
| > | I don't know who the ``you'' is in ``if you do that.''  Or else I
| > | don't know what the ``that'' is.
| > | 
| > | I gave two code samples, and g++ uses the same name mangling for both.
| > | Try it.  Is that a bug in g++?
| > | 
| > | Frankly, I don't see how g++ could do anything else.
| > 
| > The point is this.  "typeof" is a GNU/C++ extension.  Its use in a
| > function declaration should be mangled differently from any standard
| > C++ construction.  After all, the ABI has provided hook for vendor
| > extension. 
| 
| I can't see why in this case. 

the demangler is introducing a vendor-specific operator where there
weren't. 

-- Gaby