Best way to `tee` to stdout and to a process

Question

I want to do something like this:

echo 'foo' | tee /dev/stdout > >(cat)

where the stdout from echo gets sent to the terminal and to the cat process. Is there a simpler way to do this?

When I run this:

 echo 'foo' | tee >(echo 'bar')

for some reason, it does not echo 'foo' it only echoes 'bar', why?

Answer 1

You do it exactly the way you have shown:

somecommand | tee >(othercommand)

The output of somecommand would be written to the input of othercommand and to standard output.

The issue with your echo 'bar' process substitution is that it doesn't care about the input that comes via tee from echo 'foo', so it just outputs bar as quickly as it can and terminates. The tee utility then tries to write to it, but fails and therefore terminates (from receiving a PIPE signal) before it writes the string to standard output. Or, tee may have time to write the data to the process substitution, in which case both bar and foo would be printed on standard output, it's not deterministic.

You need to make sure that the command in the process substitution actually reads the data sent to it (otherwise, what would be the point of sending it data?) As Uncle Billy suggests in comments, this is easily arranged in your example by letting the process substitution simply use cat >/dev/null (assuming you're not interested in the data coming from tee):

echo 'foo' | tee >(cat >/dev/null; echo 'bar')

or

echo 'foo' | tee >(echo 'bar'; cat >/dev/null)

(these two variations would vary only in the order of the final output of the two strings)

Answer 2

The accepted answer wasn't very clear to me, as the result is piped to /dev/null, obscuring if the pipe worked.

To practice confirming the value is passed, you can cat or use tr on the piped value like this:

% echo foo | tee >(tr -d '\n') ; echo bar
foo
foobar

Answer 3

echo 'foo' | tee /dev/tty |other-program