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Question

Problem with temporary unnamed pipe in bash script

Answer 1

+2 A:

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).

codaddict 2010-02-11 13:15:36

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

sugo 2010-02-11 13:30:47

@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 2010-02-12 13:55:32

Answer 2

A:

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

soulmerge 2010-02-11 13:15:56

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.

Charles Stewart 2010-02-11 13:22:55

Answer 3

A:

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".

Charles Stewart 2010-02-11 13:27:38

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

sugo 2010-02-11 13:43:58

@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`.

Dennis Williamson 2010-02-11 15:23:45

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Problem with temporary unnamed pipe in bash script

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