Strange behaviour of command substitution in bash

Question

Command substitution:

var=$(cat /some/file.txt)

assigns output of cat command to var variable (without printing cat command output to console). Next I can print value of var variable:

echo "$var"

But

var=$(java -version)

or

var=$(fish -v)

will immediately print output of the command to console (even without echo command). Why?

Why var variable has no value now?

How can I assign output of the command (e.g. java -version) to variable?

You have a UUOC (unnecessary use of cat) in var=$(cat /some/file.txt). No need to call cat, simply var=$(</some/file.txt). — David C. Rankin
– David C. Rankin, Commented Feb 26, 2016 at 20:31

mklement0 · Accepted Answer · 2016-02-26 18:43:42Z

4

Command substitutions only capture stdout output.

Presumably your commands output to stderr.

Using output redirection, you can capture stderr as well:

var=$(java -version 2>&1) # captures both stdout and stderr

answered Feb 26, 2016 at 18:43

mklement0

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2 Comments

I agree that version information should print to stdout, but some utilities do it their own way, unfortunately.

If your requirements are more complex than this, perhaps see also stackoverflow.com/questions/962255/…