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Why is "echo foo | read a ; echo $a" not working as expected?

Answer 1

+8 A:

Due to the piping the read is executed in its own subshell.

echo foo | while read a; do echo $a; done

will do what you expect it to.

Bombe 2008-12-17 14:31:10

You know, I had that exact problem, came up with that exact response (a while loop that only executed once) and never realized why it worked. Thanks.

Paul Tomblin 2008-12-17 14:42:49

I still can't use it, because $a looses its value outside the while block!echo foo | while read a; do echo $a; done ; echo $a

Diomidis Spinellis 2008-12-17 14:54:13

@Diomidis, So put whatever you need to do with $a in the loop.

Paul Tomblin 2008-12-17 14:59:16

Yeah, I know I can do that, but this distorts the code, and it doesn't scale. I posted another answer of an approach, which I ended up using.

Diomidis Spinellis 2008-12-17 15:02:38

You should have told us more about your problem. You original question was just “why doesn’t this work” (and has been answered) but we could probably have helped you a lot better if we knew more about your problem.

Bombe 2008-12-17 15:21:10

You're right. It is difficult to find the balance between being too general and too specific.

Diomidis Spinellis 2008-12-17 20:48:01

echo foo | { read a; echo $a; } does the same thing without while.

hop 2008-12-25 13:39:31

Answer 2

+4 A:

alternative:

echo foo | (read a ; echo $a)

Edit:

If you need $a outside the subshell, you have to reverse the commands:

read a < <(echo foo); echo $a

this way the read is executed in the current process

Wimmel 2008-12-17 14:37:19

This however doesn't allow me to use the value outside the subshell.

Diomidis Spinellis 2008-12-17 14:41:39

Nice, but note that <() is not POSIX syntax.

daf 2009-12-31 00:22:18

Answer 3

A:

read expects input on a new line

while read a; do echo $a; done
foo
bar
^D

is more or less what you wanted

James Anderson 2008-12-17 14:37:35

Answer 4

A:

According to Kernighan and Pike's The Unix Programming Environment (p. 159) "none of the shell built-in commands (as opposed to the control flow primitives, like for) can be redirected with > and <", and "this might be described as a bug in the shell". This seems to explain a) why code like

ls *.c |
while read file
do
   echo $a
done

invariably works without a problem, and b) the inconsistency that redirecting from a file works.

Diomidis Spinellis 2008-12-17 14:37:47

no, it doesn't explain it. you problem is simply a scoping issue.

hop 2008-12-25 13:41:41

Would you care to explain the scopes involved in an answer?

Diomidis Spinellis 2008-12-26 12:33:09

in bash, the pipe causes the following command to be executed in a subshell. the environment variable you set with read is destroyed as soon as this subshell exits (at the ; between read and echo) and thus is not available to the next command.

hop 2008-12-30 18:47:18

Bombe already explained this in his answer.

hop 2008-12-30 18:47:49

That's a good explanation, if changed into "the pipe causes the following *built-in* command to be executed in a subshell". (External commands that are part of a pipe are definitely not run in a subshell.)

Diomidis Spinellis 2009-01-02 12:30:32

moot point. show me the non-builtin command that can change the environment for any following commands. read _has_ to be built in

hop 2009-01-03 00:08:28

btw, the man page states: "Each command in a pipeline is executed as a separate process (i.e., in a subshell)." Nothing about a distinction between built-in or not built-in there.

hop 2009-01-03 00:10:14

You're right about non-builtin commands changing the environment, and also right about the man page; I was not aware of this, I thought commands in the pipeline were spawned off directly from the parent shell.

Diomidis Spinellis 2009-01-04 19:38:24

Answer 5

+3 A:

I don't think this one has already been given:

a=`echo foo`
echo $a

Arthur Reutenauer 2008-12-17 14:41:25

This works only for reading a single value.

Diomidis Spinellis 2008-12-17 14:57:04

Sure, if you want to read several values you have to do as Bombe suggests.

Arthur Reutenauer 2008-12-17 15:08:45

Answer 6

A:

I came up with a solution that doesn't hide the variable values in a subshell, and that can also work with multiple values.

set `echo foo bar`
A=$1
B=$2

Diomidis Spinellis 2008-12-17 15:00:13

If you're going to use 'set', always use 'set --' to ensure that '-foo' doesn't screw you up.

Jonathan Leffler 2008-12-17 15:35:08

Answer 7

A:

Just FYI; in ksh it is working as expected; See also http://kornshell.com/doc/faq.html, Q13

Wimmel 2008-12-17 15:19:39

It also works in zsh.

Jonas Due Vesterheden 2008-12-20 11:11:15

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Why is "echo foo | read a ; echo $a" not working as expected?

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