Re: incorrect brace expansion when using default values
Tatavarty Kalyan schrieb am 08.09.2006 um 11:44:47 (+0800):
It is because the string
a{b,c}
is outside of the quotes. So the brace expansion comes first and
duplicates
the arguments to the echo call.
Yes, as you said the brace expansion is outside the double quotes so
shouldn't it be more like:
$ foo() { echo ${1:-ab} ${1:-ac} ; }
Yes, of course. ;)
It so simple, just eval the braces and nothing else.
I did one step too much and discovered that ab is atomar and underlies no
further evaluation.
So my assumption is correct for this example only, but not for any possible
string replacing ab as yours.
Alexander
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Re: incorrect brace expansion when using default values
Chris F.A. Johnson wrote:
Quote them, and they do expand:
$ foo() { echo ${1:-a{b,c}} ; }
$ foo
ab ac
Brace expansion is essentially separate from the rest of the expansions:
in fact, it's designed to be part of a separate library if desired. As
such, it doesn't implement all of the shell's somewhat quirky quoting
semantics. It understands simple single- and double-quoted strings and
backslash escapes. The manual is perhaps not as clear about this as it
could be.
The quoted string, once the brace expansion code finishes, comes out like
this:
${1:-ab} ${1:-ac}
From there, the results observed should be understandable.
Chet
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Re: incorrect brace expansion when using default values
Tatavarty Kalyan schrieb am 07.09.2006 um 14:20:43 (+0800):
On 9/6/06, Chris F.A. Johnson [EMAIL PROTECTED] wrote:
On 2006-09-06, Andreas Schwab wrote:
[EMAIL PROTECTED] (Paul Jarc) writes:
Mike Frysinger [EMAIL PROTECTED] wrote:
this little bit of code doesnt work right:
foo() { echo ${1:-a{b,c}} ; }
The first '}' is interpreted as the end of the parameter expansion.
It depends on the number of parameters, if the first or second brace is
taken. So this is very likely a Bug. ;)
That means in the case
foo a
the first brace is taken.
And in the case
foo
the second one does the termination.
Quote them, and they do expand:
$ foo() { echo ${1:-a{b,c}} ; }
$ foo
ab ac
However, there is a problem:
$ foo 1
1 1
Where is the second '1' coming from?
It seems
foo() { echo ${1:-a{b,c}} ; } expands to
foo() { echo ${1:-ab} ${1:-ac} ; }
Not exactly, it is acting like:
foo() { echo ${1:-ab} ${1:-ac} ; }
To check the first assumption, that there are two parameters to echo, use the
call without any parameters to foo:
$ foo
ab ac
$
The show the second, apply a string with spaces:
$ foo() { echo ${1:-a{b,c}} ; }
$ foo x
x x
$
It is because the string
a{b,c}
is outside of the quotes. So the brace expansion comes first and duplicates
the arguments to the echo call.
Alexander Elgert
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Re: incorrect brace expansion when using default values
On 9/8/06, Alexander Elgert [EMAIL PROTECTED] wrote:
Tatavarty Kalyan schrieb am 07.09.2006 um 14:20:43 (+0800):
On 9/6/06, Chris F.A. Johnson [EMAIL PROTECTED] wrote:
On 2006-09-06, Andreas Schwab wrote:
[EMAIL PROTECTED] (Paul Jarc) writes:
Mike Frysinger [EMAIL PROTECTED] wrote:
this little bit of code doesnt work right:
foo() { echo ${1:-a{b,c}} ; }
The first '}' is interpreted as the end of the parameter expansion.
It depends on the number of parameters, if the first or second brace is
taken. So this is very likely a Bug. ;)
That means in the case
foo a
the first brace is taken.
And in the case
foo
the second one does the termination.
Quote them, and they do expand:
$ foo() { echo ${1:-a{b,c}} ; }
$ foo
ab ac
However, there is a problem:
$ foo 1
1 1
Where is the second '1' coming from?
It seems
foo() { echo ${1:-a{b,c}} ; } expands to
foo() { echo ${1:-ab} ${1:-ac} ; }
Not exactly, it is acting like:
foo() { echo ${1:-ab} ${1:-ac} ; }
To check the first assumption, that there are two parameters to echo, use
the
call without any parameters to foo:
$ foo
ab ac
$
The show the second, apply a string with spaces:
$ foo() { echo ${1:-a{b,c}} ; }
$ foo x
x x
$
It is because the string
a{b,c}
is outside of the quotes. So the brace expansion comes first and
duplicates
the arguments to the echo call.
Yes, as you said the brace expansion is outside the double quotes so
shouldn't it be more like:
$ foo() { echo ${1:-ab} ${1:-ac} ; }
For example,
$ foo() { echo ${1:-a{\'b,c}} ; }
$foo
a'b ac
$foo() { echo ${1:-a\'b} ${1:-ac} ; }
$foo
a'b ac
Alexander Elgert
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Re: incorrect brace expansion when using default values
[EMAIL PROTECTED] (Paul Jarc) writes:
Mike Frysinger [EMAIL PROTECTED] wrote:
this little bit of code doesnt work right:
foo() { echo ${1:-a{b,c}} ; }
Brace expansion happens before parameter expansion (man bash,
EXPANSION).
Brace expansion doesn't come into play here, because the braces are
quoted.
Andreas.
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And now for something completely different.
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Re: incorrect brace expansion when using default values
On Wednesday 06 September 2006 05:04, Andreas Schwab wrote:
[EMAIL PROTECTED] (Paul Jarc) writes:
Mike Frysinger [EMAIL PROTECTED] wrote:
this little bit of code doesnt work right:
foo() { echo ${1:-a{b,c}} ; }
Brace expansion happens before parameter expansion (man bash,
EXPANSION).
Brace expansion doesn't come into play here, because the braces are
quoted.
without quotes has the same behavior
foo() { echo ${1:-a{b,c}} ; }
-mike
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Re: incorrect brace expansion when using default values
On 2006-09-06, Andreas Schwab wrote:
[EMAIL PROTECTED] (Paul Jarc) writes:
Mike Frysinger [EMAIL PROTECTED] wrote:
this little bit of code doesnt work right:
foo() { echo ${1:-a{b,c}} ; }
The first '}' is interpreted as the end of the parameter expansion.
Brace expansion happens before parameter expansion (man bash,
EXPANSION).
Brace expansion doesn't come into play here, because the braces are
quoted.
Quote them, and they do expand:
$ foo() { echo ${1:-a{b,c}} ; }
$ foo
ab ac
However, there is a problem:
$ foo 1
1 1
Where is the second '1' coming from?
--
Chris F.A. Johnson http://cfaj.freeshell.org
===
Author:
Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
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Re: incorrect brace expansion when using default values
Mike Frysinger [EMAIL PROTECTED] wrote:
this little bit of code doesnt work right:
foo() { echo ${1:-a{b,c}} ; }
Brace expansion happens before parameter expansion (man bash,
EXPANSION). So the first } ends the parameter expression, and the
second } isn't special. The result of parameter expansion is not
subject to brace expansion.
$ foo
a{b,c}
This is correct. a{b,c is the default value, and } follows the
parameter expression.
$ foo 1
a}
I get 1}.
paul
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