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Bash parameter expansion can't match spaces in variables like $* and $@

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1

Using dash (0.5.10.2), I can do this:

% dash
$ set -- x hello world
$ echo "<$*#x >"
<hello world>

This is the behavior I expect. The contents of $* (which are x hello world as assigned by set and delimited by spaces) are run through shell parameter expansion to remove any leading x  for echo, thus resulting in hello world, which I'm echoing with surrounding brackets to demonstrate the lack of surrounding white space.

I can't replicate that in bash (5.0.2(1)-release). It appears the space, a delimiter, is inaccessible:

% bash
$ set -- x hello world
$ echo "<$*#x >"
<x hello world>
$ echo "<$@#x >" # trying $@ instead of $*
<x hello world>
$ echo "<$*#x>" # without the space works but now I have a space
< hello world>
$ echo "<$*#x?>" # trying the `?` wildcard for a single character
<x hello world>
$ echo "<$*#x >" # trying to escape the space
<x hello world>
$ echo "<$*/#x />" # using bash pattern substitution instead
<x hello world>
$ echo "<$*#x$IFS>" # trying the input field separator variable
<x hello world>

Is there a solution here? Perhaps some way of modifying $* or changing the output field separator?

My current workaround is to assign it to a temporary variable, but that's pretty ugly. (I need bash or else I'd stick with /bin/sh, which is dash.)

bash shell parameter-expansion

share|improve this question

edited Mar 8 at 22:42

Adam Katz

asked Mar 8 at 22:18

Adam KatzAdam Katz

6,22713756

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  As I read pubs.opengroup.org/onlinepubs/9699919799/utilities/…: If parameter is #, *, or @, the result of the expansion is unspecified. -- so both are equally correct, and a script that wants to work on all POSIX platforms shouldn't rely on either behavior. (var=$* and $var// /<SPACE>, and you're set).
  
  – Charles Duffy
  Mar 8 at 22:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @CharlesDuffy - Err... Um.. isn't $var// /<SPACE> parameter expansion with substring replacement .. a bashism? POSIX only provides suffix and prefix removal as I read it.
  
  – David C. Rankin
  Mar 8 at 23:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DavidC.Rankin, quite right. Still, the point that var=$* avoids relying on undefined behavior stands.
  
  – Charles Duffy
  Mar 8 at 23:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DavidC.Rankin – This is a bash question, so the bashism is okay ... though I was using var="$*"; echo "$var#x " to the same effect because I so rarely script in bash and I try to limit my bash scripts' bashisms to where they're absolutely necessary.
  
  – Adam Katz
  Mar 8 at 23:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Agreed, I was just confused by your apparent use of dash.
  
  – David C. Rankin
  Mar 8 at 23:15
  
  

  
  
  
  

 | 
show 4 more comments

1

Using dash (0.5.10.2), I can do this:

% dash
$ set -- x hello world
$ echo "<$*#x >"
<hello world>

This is the behavior I expect. The contents of $* (which are x hello world as assigned by set and delimited by spaces) are run through shell parameter expansion to remove any leading x  for echo, thus resulting in hello world, which I'm echoing with surrounding brackets to demonstrate the lack of surrounding white space.

I can't replicate that in bash (5.0.2(1)-release). It appears the space, a delimiter, is inaccessible:

% bash
$ set -- x hello world
$ echo "<$*#x >"
<x hello world>
$ echo "<$@#x >" # trying $@ instead of $*
<x hello world>
$ echo "<$*#x>" # without the space works but now I have a space
< hello world>
$ echo "<$*#x?>" # trying the `?` wildcard for a single character
<x hello world>
$ echo "<$*#x >" # trying to escape the space
<x hello world>
$ echo "<$*/#x />" # using bash pattern substitution instead
<x hello world>
$ echo "<$*#x$IFS>" # trying the input field separator variable
<x hello world>

Is there a solution here? Perhaps some way of modifying $* or changing the output field separator?

My current workaround is to assign it to a temporary variable, but that's pretty ugly. (I need bash or else I'd stick with /bin/sh, which is dash.)

bash shell parameter-expansion

share|improve this question

edited Mar 8 at 22:42

Adam Katz

asked Mar 8 at 22:18

Adam KatzAdam Katz

6,22713756

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  As I read pubs.opengroup.org/onlinepubs/9699919799/utilities/…: If parameter is #, *, or @, the result of the expansion is unspecified. -- so both are equally correct, and a script that wants to work on all POSIX platforms shouldn't rely on either behavior. (var=$* and $var// /<SPACE>, and you're set).
  
  – Charles Duffy
  Mar 8 at 22:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @CharlesDuffy - Err... Um.. isn't $var// /<SPACE> parameter expansion with substring replacement .. a bashism? POSIX only provides suffix and prefix removal as I read it.
  
  – David C. Rankin
  Mar 8 at 23:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DavidC.Rankin, quite right. Still, the point that var=$* avoids relying on undefined behavior stands.
  
  – Charles Duffy
  Mar 8 at 23:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DavidC.Rankin – This is a bash question, so the bashism is okay ... though I was using var="$*"; echo "$var#x " to the same effect because I so rarely script in bash and I try to limit my bash scripts' bashisms to where they're absolutely necessary.
  
  – Adam Katz
  Mar 8 at 23:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Agreed, I was just confused by your apparent use of dash.
  
  – David C. Rankin
  Mar 8 at 23:15
  
  

  
  
  
  

 | 
show 4 more comments

Using dash (0.5.10.2), I can do this:

% dash
$ set -- x hello world
$ echo "<$*#x >"
<hello world>

This is the behavior I expect. The contents of $* (which are x hello world as assigned by set and delimited by spaces) are run through shell parameter expansion to remove any leading x  for echo, thus resulting in hello world, which I'm echoing with surrounding brackets to demonstrate the lack of surrounding white space.

I can't replicate that in bash (5.0.2(1)-release). It appears the space, a delimiter, is inaccessible:

% bash
$ set -- x hello world
$ echo "<$*#x >"
<x hello world>
$ echo "<$@#x >" # trying $@ instead of $*
<x hello world>
$ echo "<$*#x>" # without the space works but now I have a space
< hello world>
$ echo "<$*#x?>" # trying the `?` wildcard for a single character
<x hello world>
$ echo "<$*#x >" # trying to escape the space
<x hello world>
$ echo "<$*/#x />" # using bash pattern substitution instead
<x hello world>
$ echo "<$*#x$IFS>" # trying the input field separator variable
<x hello world>

Is there a solution here? Perhaps some way of modifying $* or changing the output field separator?

My current workaround is to assign it to a temporary variable, but that's pretty ugly. (I need bash or else I'd stick with /bin/sh, which is dash.)

bash shell parameter-expansion

share|improve this question

edited Mar 8 at 22:42

Adam Katz

asked Mar 8 at 22:18

Adam KatzAdam Katz

6,22713756

% dash
$ set -- x hello world
$ echo "<$*#x >"
<hello world>

% bash
$ set -- x hello world
$ echo "<$*#x >"
<x hello world>
$ echo "<$@#x >" # trying $@ instead of $*
<x hello world>
$ echo "<$*#x>" # without the space works but now I have a space
< hello world>
$ echo "<$*#x?>" # trying the `?` wildcard for a single character
<x hello world>
$ echo "<$*#x >" # trying to escape the space
<x hello world>
$ echo "<$*/#x />" # using bash pattern substitution instead
<x hello world>
$ echo "<$*#x$IFS>" # trying the input field separator variable
<x hello world>

share|improve this question

edited Mar 8 at 22:42

Adam Katz

asked Mar 8 at 22:18

Adam KatzAdam Katz

6,22713756

share|improve this question

edited Mar 8 at 22:42

Adam Katz

asked Mar 8 at 22:18

Adam KatzAdam Katz

6,22713756

asked Mar 8 at 22:18

Adam KatzAdam Katz

6,22713756

asked Mar 8 at 22:18

Adam KatzAdam Katz

6,22713756

1 Answer
1

active

oldest

votes

2

For array-like operands, the string operation is applied for each element before they are joined on spaces. Therefore, you can't apply them to the joining space.

Here's an example showing this:

$ set -- hello world "hello world with spaces"
$ echo "$*// /<SPACE>"
hello world hello<SPACE>world<SPACE>with<SPACE>spaces

Spaces within each argument is replaced just fine, but the spaces between them as inserted by $* are not affected.

The workaround is indeed a temporary variable.

share|improve this answer

edited Mar 8 at 22:32

answered Mar 8 at 22:26

that other guythat other guy

75k886124

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ah, nice demonstration. echo "<$*#hello >" (using your set line) better fits what I'm doing, and sure enough it gives me <hello world world with spaces> because of the implicit for each that you noted.
  
  – Adam Katz
  Mar 8 at 22:37
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  If your real question is "How do I concatenate all arguments except the first one?" then the answer is "$*:2"
  
  – that other guy
  Mar 8 at 22:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Actually, yes: that's the literal answer to my (oversimplified) question, though I think your first paragraph and Charles Duffy's comment to the question combine for a more informative answer. (Also, I had a whole bunch of conditionals like [ "$*" != "$*#* foo* bar" ] which I've had to convert to case patterns like ( * foo* bar* ) but that didn't make it to the simplified question I asked here.)
  
  – Adam Katz
  Mar 8 at 22:50
  
  

  
  
  
  

add a comment | 

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2

For array-like operands, the string operation is applied for each element before they are joined on spaces. Therefore, you can't apply them to the joining space.

Here's an example showing this:

$ set -- hello world "hello world with spaces"
$ echo "$*// /<SPACE>"
hello world hello<SPACE>world<SPACE>with<SPACE>spaces

Spaces within each argument is replaced just fine, but the spaces between them as inserted by $* are not affected.

The workaround is indeed a temporary variable.

share|improve this answer

edited Mar 8 at 22:32

answered Mar 8 at 22:26

that other guythat other guy

75k886124

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ah, nice demonstration. echo "<$*#hello >" (using your set line) better fits what I'm doing, and sure enough it gives me <hello world world with spaces> because of the implicit for each that you noted.
  
  – Adam Katz
  Mar 8 at 22:37
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  If your real question is "How do I concatenate all arguments except the first one?" then the answer is "$*:2"
  
  – that other guy
  Mar 8 at 22:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Actually, yes: that's the literal answer to my (oversimplified) question, though I think your first paragraph and Charles Duffy's comment to the question combine for a more informative answer. (Also, I had a whole bunch of conditionals like [ "$*" != "$*#* foo* bar" ] which I've had to convert to case patterns like ( * foo* bar* ) but that didn't make it to the simplified question I asked here.)
  
  – Adam Katz
  Mar 8 at 22:50
  
  

  
  
  
  

add a comment | 

2

For array-like operands, the string operation is applied for each element before they are joined on spaces. Therefore, you can't apply them to the joining space.

Here's an example showing this:

$ set -- hello world "hello world with spaces"
$ echo "$*// /<SPACE>"
hello world hello<SPACE>world<SPACE>with<SPACE>spaces

Spaces within each argument is replaced just fine, but the spaces between them as inserted by $* are not affected.

The workaround is indeed a temporary variable.

share|improve this answer

edited Mar 8 at 22:32

answered Mar 8 at 22:26

that other guythat other guy

75k886124

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ah, nice demonstration. echo "<$*#hello >" (using your set line) better fits what I'm doing, and sure enough it gives me <hello world world with spaces> because of the implicit for each that you noted.
  
  – Adam Katz
  Mar 8 at 22:37
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  If your real question is "How do I concatenate all arguments except the first one?" then the answer is "$*:2"
  
  – that other guy
  Mar 8 at 22:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Actually, yes: that's the literal answer to my (oversimplified) question, though I think your first paragraph and Charles Duffy's comment to the question combine for a more informative answer. (Also, I had a whole bunch of conditionals like [ "$*" != "$*#* foo* bar" ] which I've had to convert to case patterns like ( * foo* bar* ) but that didn't make it to the simplified question I asked here.)
  
  – Adam Katz
  Mar 8 at 22:50
  
  

  
  
  
  

add a comment | 

For array-like operands, the string operation is applied for each element before they are joined on spaces. Therefore, you can't apply them to the joining space.

Here's an example showing this:

$ set -- hello world "hello world with spaces"
$ echo "$*// /<SPACE>"
hello world hello<SPACE>world<SPACE>with<SPACE>spaces

Spaces within each argument is replaced just fine, but the spaces between them as inserted by $* are not affected.

The workaround is indeed a temporary variable.

share|improve this answer

edited Mar 8 at 22:32

answered Mar 8 at 22:26

that other guythat other guy

75k886124

$ set -- hello world "hello world with spaces"
$ echo "$*// /<SPACE>"
hello world hello<SPACE>world<SPACE>with<SPACE>spaces

share|improve this answer

edited Mar 8 at 22:32

answered Mar 8 at 22:26

that other guythat other guy

75k886124

answered Mar 8 at 22:26

that other guythat other guy

75k886124

answered Mar 8 at 22:26

that other guythat other guy

75k886124