Bug#259196: marked as done (libstdc++5: mixed signed/unsigned char comparison in std::char_traits<char>)
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and subject line Bug#259196: libstdc++5: mixed signed/unsigned char comparison in std::char_traits<char>
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Received: (at submit) by bugs.debian.org; 13 Jul 2004 11:51:47 +0000
>From giuseppe@bohr.pisa.iol.it Tue Jul 13 04:51:47 2004
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From: giuseppe bonacci <g.bonacci@libero.it>
To: Debian Bug Tracking System <submit@bugs.debian.org>
Subject: libstdc++5: mixed signed/unsigned char comparison in std::char_traits<char>
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Date: Tue, 13 Jul 2004 13:51:10 +0200
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Package: libstdc++5
Version: 1:3.3.4-2
Severity: minor
I report this as "minor" since I don't know if it's really a bug or
standard behaviour that I'm not understanding.
The following program (to be compiled with g++ 3.x)
#include <iostream>
#include <string>
#define BLURB(x) #x << "\t=3D=3D " << (x) << '\n'
int main()
{
std::basic_string<char> s1, s2;
s1 =3D 0xe0;
s2 =3D 'a';
std::cout << "s1 =3D=3D '" << s1 << "'\ns2 =3D=3D '" << s2 << "'\n";
std::cout << BLURB( (s1 < s2) );
std::cout << BLURB( (s1[0] < s2[0]) );
std::cout << BLURB( std::char_traits<char>::lt(s1[0], s2[0]) );
std::cout << BLURB( std::char_traits<char>::compare(s1.c_str(),
s2.c_str(), 1) );
std::cout << BLURB( s1.compare(s2) );
return 0;
}
produces this output on my x86 linux pc:
s1 =3D=3D '=E0'
s2 =3D=3D 'a'
(s1 < s2) =3D=3D 0
(s1[0] < s2[0]) =3D=3D 1
std::char_traits<char>::lt(s1[0], s2[0]) =3D=3D 1
std::char_traits<char>::compare(s1.c_str(), s2.c_str(), 1) =3D=3D 1
s1.compare(s2) =3D=3D 1
The above results are counter-intuitive, but agree with the
behaviour of C standard library (1999 standard):=20
- strcmp and memcmp treat their arguments as unsigned char*, so
that strcmp(s1.c_str(), s2.c_str()) > 0 (meaning s1 > s2)
- on my platform char is signed, so s1[0] < s2[0] (because s1[0] < 0)
On the other hand, Stroustrup's TC++PL, section 20.2.1 "Character traits"
reports "The compare() function uses lt() and eq() to compare characters.=
",=20
so I'd expect
std::char_traits<char>::lt(s1[0], s2[0])
and=20
std::char_traits<char>::compare(s1.c_str(), s2.c_str(), 1)
to return consistent results, which they do not. As a side effect,
s1.compare(s2) is not consistent with std::char_traits<char>::lt()
either.
As far as I can see, GNU libstdc++5 implementation of=20
std::char_traits<char>::compare() uses memcmp() instead of lt(), and so
inherits the unsigned char comparison, while std::char_traits<char>::lt()
plainly uses '<' to compare its arguments, keeping them signed.
It's quite likely that I am missing something in Stroustrup's book.
What does the standard mandate?
Best regards
giuseppe
-- System Information:
Debian Release: testing/unstable
APT prefers testing
APT policy: (500, 'testing')
Architecture: i386 (i686)
Kernel: Linux 2.4.26-1-686
Locale: LANG=3DC, LC_CTYPE=3DC
Versions of packages libstdc++5 depends on:
ii gcc-3.3-base 1:3.3.4-2 The GNU Compiler Collection =
(base=20
ii libc6 2.3.2.ds1-13 GNU C Library: Shared librar=
ies an
ii libgcc1 1:3.3.4-2 GCC support library
-- no debconf information
---------------------------------------
Received: (at 259196-done) by bugs.debian.org; 22 May 2005 10:16:52 +0000
>From falk@debian.org Sun May 22 03:16:52 2005
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To: 259196-done@bugs.debian.org
Subject: Bug#259196: libstdc++5: mixed signed/unsigned char comparison in
std::char_traits<char>
From: Falk Hueffner <falk@debian.org>
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Date: Sun, 22 May 2005 12:16:19 +0200
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Hi,
acccording to http://gcc.gnu.org/PR15276, the behaviour has now been
codified by a DR as correct, so closing.
--
Falk
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