CODEWITHSUNDEEP
shellzsh
D0SBoots
D0SBoots

Reputation: 850

"Pipe stderr, and not stdout" not working in Zsh

How to pipe stderr, and not stdout? perfecly captures my problem, and the first answer is exactly how I initially tried to solve it:

(echo stdout; echo 1>&2 stderr) 2>&1 >/dev/null | less

(The subshell command with echos is a minimal placeholder thanks to user1934428 to demonstrate the issue; my actual problem has a more useful command, but one that everyone else can't run.)

However, it doesn't work: It's showing both stdout and stderr together. If I remove the pipe, it works as expected, only showing stderr. Eventually I realized this might be a shell thing, and tried bash: It worked perfectly there. I'm using Zsh: what about Zsh makes this fail?

Upvotes: 4

Views: 2859

Answers (2)

WolleTD
WolleTD

Reputation: 400

This is due to MULTIOS, which duplicates the output streams when you supply multiple redirections. Pipes are implicit redirections.

Type ls >/dev/null | less in your zsh and you will see output even though you redirected it. That's multios duplicating the stream.

unsetopt multios will provide the desired behaviour. This is quite a handy feature if I wouldn't always forget about it until it irritates me.

http://zsh.sourceforge.net/Doc/Release/Redirection.html#Multios

Upvotes: 8

user1934428
user1934428

Reputation: 22225

Both behave the same for me, in that they only show stderr.

For testing, I created a command which outputs to stdout and stderr, and works the same in Zsh and bash:

(echo stdout; echo 1>&2 stderr)

prints

stdout
stderr

Now I do your redirection:

 (echo stdout; echo 1>&2 stderr) 2>&1 >/dev/null

Here, I see on bash and zsh

stderr

but of course written to standard output, because 2>&1 first redirects the stderr of the command to what is at the moment standard output, and then discards the standard output of the command.

Upvotes: 1

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