the macro __FUNCTION__
returns the name of the constructor rather than the name of the destructor
in when used in the destructor.
test::test()
{
cerr << __FUNCTION__ << endl; // should output test
}
test::~test()
{
cerr << __FUNCTION__ << endl; // should output ~test?, but outputs test.
}
I'm using g++ 3.3 on MAC OS X 10.3
gcc version 3.3 20030304 (Apple Computer, Inc. build 1495) 6 1483
On Sun, 18 Apr 2004 20:01:27 GMT in comp.lang.c++, "JustSomeGuy"
<no**@nottelling.com> wrote, the macro __FUNCTION__ returns the name of the constructor rather than the name of the destructor in when used in the destructor.
If you believe that something is a compiler bug, please avoid posting it
in comp.lang.c++.
The obvious place for this would be gnu.g++.bug
JustSomeGuy wrote in news:r8Bgc.161060$Pk3.87406@pd7tw1no in
comp.lang.c++: the macro __FUNCTION__ returns the name of the constructor rather than the name of the destructor in when used in the destructor.
test::test() { cerr << __FUNCTION__ << endl; // should output test } test::~test() { cerr << __FUNCTION__ << endl; // should output ~test?, but outputs test. }
I'm using g++ 3.3 on MAC OS X 10.3 gcc version 3.3 20030304 (Apple Computer, Inc. build 1495)
Ask this in news://gnu.gcc.help , __FUNCTION__ isn't a standard
macro so it does what the gcc team say it does, and they may say
that in ctor's/dtor's it returns the class name.
Rob.
-- http://www.victim-prime.dsl.pipex.com/
Rob Williscroft <rt*@freenet.co.uk> spoke thus: Ask this in news://gnu.gcc.help , __FUNCTION__ isn't a standard macro so it does what the gcc team say it does, and they may say that in ctor's/dtor's it returns the class name.
__FUNCTION__ isn't standard, but C99 specifies the __func__ predefined
identifier, which I believe is #define'd to be __FUNCTION__ for OP's
implementation. I would be interested to know whether __func__ made
it into standard C++ as well.
--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cyberspace.org | don't, I need to know. Flames welcome.
Christopher Benson-Manica wrote: __FUNCTION__ isn't standard, but C99 specifies the __func__ predefined identifier, which I believe is #define'd to be __FUNCTION__ for OP's implementation. I would be interested to know whether __func__ made it into standard C++ as well.
Not yet. C++ hasn't added any new features since the first standard in 1998.
-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
Kevin Goodsell <us*********************@neverbox.com> spoke thus:
(moving to comp.std.c++) Christopher Benson-Manica wrote:
__FUNCTION__ isn't standard, but C99 specifies the __func__ predefined identifier, which I believe is #define'd to be __FUNCTION__ for OP's implementation. I would be interested to know whether __func__ made it into standard C++ as well.
Not yet. C++ hasn't added any new features since the first standard in 1998.
Does "not yet" mean that it's planned for the next version of the C++
standard?
--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cyberspace.org | don't, I need to know. Flames welcome.
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[ FAQ: http://www.jamesd.demon.co.uk/csc/faq.html ] at***@nospam.cyberspace.org (Christopher Benson-Manica) wrote in message news:<c6**********@chessie.cirr.com>... Kevin Goodsell <us*********************@neverbox.com> spoke thus:
(moving to comp.std.c++)
Christopher Benson-Manica wrote:
__FUNCTION__ isn't standard, but C99 specifies the __func__ predefined identifier, which I believe is #define'd to be __FUNCTION__ for OP's implementation. I would be interested to know whether __func__ made it into standard C++ as well.
Not yet. C++ hasn't added any new features since the first standard in 1998.
Does "not yet" mean that it's planned for the next version of the C++ standard?
There is some obvious discussion about the meaning of __func__ in C++
special cases - the destructor mentioned before, namespaces - but it is
covered by the desire to create a viable subset of C99 and C++0x. So
s/planned/considered/ and you're right.
Regards,
Michiel Salters
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