Feed source command with a pipe

Question

Previously I used source command like this:

source file_name

But what I'm trying to do is this:

echo something | source

Which doesn't work.

score 70 · Accepted Answer · edited Apr 01 '20 at 13:16

70

Since source (or .) takes a file as argument, you could try process substitution:

source <(echo something)

edited Apr 01 '20 at 13:16

nnnmmm

155

answered Dec 31 '14 at 20:50

muru

72,889

1

That worked, thank you. And Happy New Year! – Desmond Hume Dec 31 '14 at 20:54
1

@muru This works great. Note that "| source /dev/stdin" has the appearance of working, but it'll actually drop all of the variable assignments. Not sure why. Seems solid in theory. It's insanity. – Dustin Oprea Nov 17 '17 at 03:25
8

@DustinOprea pipes create a subshell, so all those variable assignments never affect the parent shell. – muru Nov 17 '17 at 03:27
https://mywiki.wooledge.org/SubShell – MarcH Oct 22 '20 at 00:50
Works in Bash: source /dev/stdin <<< "echo 123; X=42; echo 456"; echo $X – Martin Feb 27 '22 at 05:45
The question is not about a specific shell so it is important to say that the process substitution <(command ...) is not part of the POSIX specification. You can find it in ksh, bash and some other shells. See What is the portable (POSIX) way to achieve process substitution?. – pabouk - Ukraine stay strong Mar 29 '22 at 07:15
@pabouk-Ukrainestaystrong eh... did you miss that source isn't part of the POSIX specification either? – muru Mar 29 '22 at 10:38
That makes it even worse :D The . command is part of POSIX and source is just a more readable alias for it. See https://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#dot – pabouk - Ukraine stay strong Mar 29 '22 at 12:01
@pabouk-Ukrainestaystrong I'm not sure how it makes it worse, since source is explicitly in the question, so this question at least isn't about POSIX shells – muru Mar 29 '22 at 12:02
I wonder why sudo cat /root/.bashrc | bash didn't work. But your command worked perfectly. – Shayan Mar 02 '23 at 10:04

mikeserv · Answer 2 · 2014-12-31T21:09:22.767

Your source command requires a file argument. You can get that in some shells with a process substitution, and this is because in the same way the shell replaces...

arg=$(echo hi)

...the echo bit there on the command-line with the subshell's output, in the case of process substitution it replaces the subshell with a named file - usually /dev/fd/62 or something - some link to a file-descriptor. With a pipe the file descriptor is 0 so...

echo 'echo hi' | . /dev/fd/0

... /dev/stdin or whatever as the case may be should work just fine on any linux system - and many others besides. You can also use here-documents similarly:

. /dev/fd/3 3<<HI
$(echo 'echo hi')
HI

You can verify the way your shell handles process substitution, by the way:

(set -x; readlink <(:))

...which prints (in bash):

+ set +x
+ readlink /dev/fd/63
++ :
pipe:[2212581]

...and so we can see that the shell is doing the substitution and readlink is reading from an an anoymous pipe that it opens on file-descriptor 63.

You cannot use a process in a pipe (your example echo 'echo hi' | . /dev/fd/0) to make side-effects in the current shell (like setting aliases, environment etc.) because the commands in a pipe are run as child processes and their side-effect are lost when they are finished. — pabouk - Ukraine stay strong, Mar 29 '22 at 07:20

Feed source command with a pipe

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