Problem with temporary unnamed pipe in bash script

Question

I have the following question:

When I execute the following script directly in a terminal window, the commands behave as expected.

$ diff <(echo tmp) <(echo tmp1)
1c1
< tmp
---
> tmp1

However when I write the same command in a shell script

#! /bin/bash
diff <(echo tmp) <(echo tmp1)

I get the following error message:

$ sh test.sh
test.sh: line 2: syntax error near unexpected token `('
test.sh: line 2: ` diff <(echo tmp) <(echo tmp1)'

Initially I thought this was an issue with diff, but this also happens with other commands. Does anybody have an idea what causes the problem?

codaddict · Accepted Answer · 2010-02-11 13:15:36Z

5

Try

bash test.sh

or

chmod ugo+x test.sh
./test.sh

Works fine for me when I do either.

Looks like the syntax is not supported by the bourne shell (sh).

answered Feb 11, 2010 at 13:15

codaddict

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2 Comments

sugo Over a year ago

Thanks, did not think of the differences between bash and bourne. This solved the problem!

Charles Stewart Over a year ago

@sugo: If your problem is solved, do consider accepting codaddict's answer. There's a tick outline under the answer's score you click on to do that.

Charles Stewart · Accepted Answer · 2010-02-11 13:54:57Z

1

When bash is invoked using sh, it starts up in a special, POSIX-compliant mode. This has different syntax, which I guess explains the different results.

See bashref of POSIX mode, #22: "process substitution is not available".

edited Feb 11, 2010 at 13:54

answered Feb 11, 2010 at 13:27

Charles Stewart

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2 Comments

sugo Over a year ago

Thanks for the clarification, I though that #! /bin/bash at the beginning of the script would be sufficient.

Dennis Williamson Over a year ago

@sugo: The shebang is overridden when you specify the interpreter. It isn't necessarily running Bash in POSIX mode unless sh is symlinked to bash. So it's actually using whatever sh represents on your system. One way to see this is to create a file consisting of ls -l /proc/$$/exe called "showshell" and run it with sh ./showshell and bash ./showshell, and you'll see the symlink. On my system, sh is /bin/dash.

soulmerge · Accepted Answer · 2010-02-11 13:15:56Z

0

That syntax doesn't look familiar. Are you sure you are using bash in your terminal? You can verify by typing echo $SHELL.

answered Feb 11, 2010 at 13:15

soulmerge

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1 Comment

Charles Stewart Over a year ago

I have this working as per qn on a bash on osx, and failing for a bash on debian, both version 2.05b.0(1)-release.

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