Bash 参数扩展和数组索引操作

Question

The ${parameter,,pattern} parameter expansion converts alphabetic characters in parameter to lower-case.

On cygwin 1.7.11-1 Bash 4.1.10(4) and also on my debian squeeze Bash 4.1.5(1) ;
if i run the following, i get a curious result:

$ declare -a a=(Zero One Two Three); n=0; echo "${a[n],,}->${n}"; echo "${a[++n]}->${n}"; echo "${a[++n],,}->${n}"
zero->0
One->1
three->3
$

NB: similar results happen:
for ,,* or ^^ case conversion;
for some other expansions such as ${parameter##word} ;
for using either prefix/postfix ++ or -- operator;
for using $((++n)) instead of just ++n .

However, the length expansion ${#parameter} works as i might expect:
in the above snippet, echo "${#a[++n]}->${n}" instead of echo "${a[++n],,}->${n} would yield 3->2 instead of three->3 ~& the length of a[2]="two" is indeed 3 characters.

I imagine that the parameter expansion is happening twice. But why is this happening?

Answer 1

I grabbed the latest bash source & it appears that this issue is resolved for this version:

$ ./bash --version
GNU bash, version 4.2.24(1)-release (x86_64-unknown-linux-gnu)
Copyright (C) 2011 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
$

Note however that on my Debian Squeeze stable , I have:

$ apt-cache policy bash
bash:
  Installed: 4.1-3
  Candidate: 4.1-3
  Version table:
 *** 4.1-3 0
        500 http://ftp.uk.debian.org/debian/ squeeze/main amd64 Packages
        100 /var/lib/dpkg/status
$
$ bash --version
GNU bash, version 4.1.5(1)-release (x86_64-pc-linux-gnu)
Copyright (C) 2009 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

I think this means that currently on Debian Squeeze , the latest stable version of bash is 4.1.5(1) & contains this bug. Also on Cygwin 1.7.11-1 currently the latest available bash package appears to be version 4.1.10(4) & contains this bug. Of course the issue is easy to work around, so no real need to build the 4.2 sources.

Answer 2

I got

Zero->0
One->1
Two->2

However I think "${a[++n]}->${n}" may not be well defined.

try substitute with

let n=$n+1 ; echo "${a[n],,}->${n}";

To inject the notion of sequence.