gistfile1.txt

#!/bin/bash

for name in $(cat repos.txt); do
    echo $name;
    repo=$(echo $name | cut -d'/' -f2);
    echo $repo;
    hub clone $name;
    cd $repo;
    hub fork;
    pwd && grep -lZR 'panic!' . | xargs -0 -l sed -i -e 's/panic!/panic!/g';
    git add -A;
    git commit -m 'fail -> panic';
    hub pull-request -m <<-EOM fail -> panic
			
			Sorry I broke your code with rust-lang/rust#17894 ! Here's a fix :heart:
			
			(this is a semi-automatic PR, so sorry if it's not perfect. Let me know and I'll fix any problems.)'
EOM
    cd ..;
done

sed -i'' for in place?

That might be for OS X? I don't remember the difference with the GNU version. Mind you the -i'' means no backups, before you try it.

The sed works when i type it in a regular shell.

Yep. -i -e will suffix the modified files with -e, which means that it doesn't recognise -e as a switch (just the value of -i. That should fix it.

Ah, then that's a GNU vs BSD thing.

I find in bash I always have issues using ! without escaping it on the terminal line. Is it possible your xargs is getting tripped up on the ! ?

@steveklabnik - when you say the sed works fine, are you talking about the sed part only or the whole of line 10? I'm wondering if it's xargs failing and/or quoting issues inside a bash script vs. command line (as @mgpcoe notes above)

I ran the script with set -e (to stop on error) and set -x (to show output as executed).

The relevant bit of output (OS X):

+ grep -lZR 'panic!' .
+ xargs -0 -l sed -i -e 's/panic!/panic!/g'
xargs: illegal option -- l
usage: xargs [-0opt] [-E eofstr] [-I replstr [-R replacements]] [-J replstr]
             [-L number] [-n number [-x]] [-P maxprocs] [-s size]
             [utility [argument ...]]

My xargs (OS X) doesn't have a -l (and the GNU man page says -l is deprecated anyway). I switched it out for -L 1 and that line completes (and instead has problems with "nothing to commit" after replacing panic with panic.