Pipe input consumed where I didn't expect it to

Pipe input consumed where I didn't expect it to

[Jesper Nygårds]

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I recently tried a command line of various pipes leading in to a shell
script that I've written, and it didn't behave as I expected it to do.

This reduced example demonstrates what I did. Suppose I have the following
script called streamtest:

#!/bin/zsh
if [[ ! -t 0 ]]; then
    while read x; do print "x $x"; done
else
    print "no pipe!"
fi

If I call it like this:
% for c in 1 2 3; do ./streamtest; done

I get the output:
no pipe!
no pipe!
no pipe!

No surprise there.

However, if I call it like this:
% for c in 1 2 3; do echo $c; done | while read f; do ./streamtest; done

I get the output:
x 2
x 3

Obviously my script consumes line 2 and 3, and prints them. Why is this the
case? I naively thought that the outer read (of f) on the command line
would be fed from the pipe, and then call "streamtest" without a pipe
leading into the script. In other words I thought I would get the output:

no pipe!
no pipe!
no pipe!

I assume this is just a misunderstand on my part, but if I wanted the
behaviour that I expected, how could I fix it?

Re: Pipe input consumed where I didn't expect it to

[Jun T.]

On 2016/12/05, at 6:13, Jesper Nygårds <jesper.nygards@gmail.com> wrote:

> However, if I call it like this:
> % for c in 1 2 3; do echo $c; done | while read f; do ./streamtest; done
> 
> I get the output:
> x 2
> x 3
> 
> Obviously my script consumes line 2 and 3, and prints them. Why is this the
> case? I naively thought that the outer read (of f) on the command line
> would be fed from the pipe, and then call "streamtest" without a pipe
> leading into the script.

When the while loop starts, 'read f' is called once and consumes 1. Since
the read succeeds, now streamtest is called, which consumes rest of the
input (2 3) and exists. Now 'read f' is called again but fails, so the
loop ends.