Bash – Linux command substitution order

bashcommand-substitutionshell

I'm confused with the command substitution. I think command substitution as like a programming language macro. The sub shell is executed first and the $(...) is substituted with it's standard output, before the parent command is evaluated. But is this all the truth?

When I try to execute the following command

echo {0..9} | xargs -n 2 $(echo 'echo | tac')

I would except to see the following output

8 9
6 7
4 5
2 3
0 1

But instead get this

| tac 0 1
| tac 2 3
| tac 4 5
| tac 6 7
| tac 8 9

So what causes | tac to be captured as an argument to the echo, and not as part of the command line? What is the exact mechanism and order, how the command line and command substitution is parsed, scoped and evaluated? Is it possible at all to build command lines dynamically in Linux in a metaprogramming style of way?

Edit

I know I could just use echo {0..9}| xargs -n 2 | tac. This question is theoretical, I'm interested why the sub shell example doesn't produce the same result

Best Answer

The genesis of the issue you are seeing is due to the way tokenization happens of a command line. In this particular case, much before the $(echo 'echo | tac') is expanded, the shell has already figured out that you intend to run a command (echo {…}) and pass it's output via a pipe (|) to another command (xargs -n 2 $(…)).

Then in the next stage does the filling in of $(…) and brace expansion {…} happens to generate the actual commands to be run. And in this stage if it results in a pipe character, so what, it's too late and it's clearly missed the bus.

Now it will be treated as not as a special metacharacter, but will be included in the xargs command line as just any ordinary (=> non-metachar) char.

Should you want to take another shot at it, then you need to eval it.

eval "echo {0..9} | xargs -n 2 $(echo 'echo | tac')"

This will output what you expect.

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