On 10/01/2012 07:56 PM, David Korn wrote:
cc: ast-developers@research.att.com
Subject: Re: [ast-developers] string <> comparison in [[ ]]
--------
one our user complains about string comparison result:
var1="4000"
var2=" 500"
[[ $var2 > $var1 ]]; echo $?
0
[[ "$var2" > "$var1" ]]; echo $?
0
Based on ksh man page:
string1 < string2
True, if string1 comes before string2 based on ASCII value of their
characters.
Space is ASCII 32, '4' is ASCII 52, thus space is less than 4. So, as
per ASCII value comparison, "4000" should have been greater than " 500",
whereas result is shows " 500" is greater.
Odd is that old bash gave "correct" result, but the new one gives the
"wrong" one - the same as ksh, so I wonder if I miss something.
Michal
I get 1 and 1 when I run this.
$ uname -a
Linux terra.research.att.com 2.6.32-279.1.1.el6.x86_64 #1 SMP Tue Jul 10
13:47:21 UTC 2012 x86_64 x86_64 i386-64 GNU/Linux
$ print $LC_ALL
C
Interesting. Is there some reason why locale affects it?
It seems to depend on encoding. For my Czech locale:
$ export LC_ALL=cs_CZ.utf-8
$ [[ "$var2" > "$var1" ]]; echo $?
0
$ export LC_ALL=cs_CZ.iso-8859-2
$ [[ "$var2" > "$var1" ]]; echo $?
0
$ export LC_ALL=cs_CZ.iso-8859-1
$ [[ "$var2" > "$var1" ]]; echo $?
1
so the result is incorrect for "usual" Czech encodings (iso-8859-2 and
utf-8 have all Czech letters, iso-8859-1 does not)
for English locale:
$ export LC_ALL=en_US.utf-8
$ [[ "$var2" > "$var1" ]]; echo $?
0
$ export LC_ALL=en_US.iso-8859-1
$ [[ "$var2" > "$var1" ]]; echo $?
0
$ export LC_ALL=en_US.iso-8859-2
$ [[ "$var2" > "$var1" ]]; echo $?
1
the result is again incorrect for usual encodings.
Could someone explain me what is happening here?
My knowledge about locales seems to be insufficient.
Thanks
Michal
David Korn
d...@research.att.com
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