Ubuntu – Why script with /bin/bash is working, with /bin/sh not

bashscripts

My friend asked me why this two simple script are not working in both shells:

Test File "abc.txt":

aaa
111
bbb
111
ccc
111
ddd

Script #1 (a):

#!/bin/sh
while read -r line ; do
    echo "Processing $line"
done < <(grep 111 abc.txt)

Output:

./a: line 4: syntax error near unexpected token `<'
./a: line 4: `done < <(grep 111 abc.txt)'

Script #2 (b):

#!/bin/bash
while read -r line ; do
    echo "Processing $line"
done < <(grep 111 abc.txt)

Output:

Processing 111
Processing 111
Processing 111

I checked bash and sh on my machine and if I understand this correct this is the same thing.

/bin/sh is just a link to /bin/bash:

-rwxr-xr-x  1 root root 938736 May 10  2012 bash
lrwxrwxrwx. 1 root root      4 Feb 13 10:20 sh -> bash

Can someone explain me where is the difference?

Best Answer

It's because Bash behaves differently when $0 is sh.

From Bash's behaviour [Bash Hackers Wiki]:

SH mode

When Bash starts in SH compatiblity mode, it tries to mimic the startup behaviour of historical versions of sh as closely as possible, while conforming to the POSIX® standard as well. The profile files read are /etc/profile and ~/.profile, if it's a login shell.

If it's not a login shell, the environment variable ENV is evaluated and the resulting filename is taken as name of the startup file.

After the startup files are read, Bash enters the POSIX(r) compatiblity mode (for running, not for starting!).

Bash starts in sh compatiblity mode when:

  • the base filename in argv[0] is sh (Attention dear uber-clever Linux users… /bin/sh may be linked to /bin/bash, but that doesn't mean it acts like /bin/bash)

More information can be found at the Bash Reference Manual: Bash POSIX Mode, specifically:

  • Process substitution is not available.

which is why your sh script is failing, as the <(..) syntax is using process substitution.

Related Question