bash – Variable Resolution in Subshell with bash -c

bash

How do I resolve a variable in a subshell inside of "bash -c"?

In the following, the second subshell — the one with an "echo" inside — resolves correctly. The first subshell — with "touch" — does not.

/bin/bash -c "\
A=/tmp/foo; \
echo $( touch \$A;  ); \
echo $( echo in subshell, \$A; ); \
"

Best Answer

It is simply, because the subshell is evaluated from your current shell and not from your subshell. Escaping the $() will make it work as you expected:

/bin/bash -c "\
A=/tmp/foo; \
echo \$( touch \$A;  ); \
echo \$( echo in subshell, \$A; ); \
"

Related Solutions

Bash – Rule for Invoking Subshell

The parentheses always start a subshell. What's happening is that bash detects that sleep 5 is the last command executed by that subshell, so it calls exec instead of fork+exec. The sleep command replaces the subshell in the same process.

In other words, the base case is:

( … ) create a subshell. The original process calls fork and wait. In the subprocess, which is a subshell:
1. sleep is an external command which requires a subprocess of the subprocess. The subshell calls fork and wait. In the subsubprocess:
  1. The subsubprocess executes the external command → exec.
  2. Eventually the command terminates → exit.
2. wait completes in the subshell.
wait completes in the original process.

The optimization is:

( … ) create a subshell. The original process calls fork and wait. In the subprocess, which is a subshell until it calls exec:
1. sleep is an external command, and it's the last thing this process needs to do.
2. The subprocess executes the external command → exec.
3. Eventually the command terminates → exit.
wait completes in the original process.

When you add something else after the call the sleep, the subshell needs to be kept around, so this optimization can't happen.

When you add something else before the call to sleep, the optimization could be made (and ksh does it), but bash doesn't do it (it's very conservative with this optimization).

Bash Shell – How to Make Explicit Operator Precedence Without Creating a Subshell

from man bash:

   { list; }
          list  is  simply executed in the current shell environment.  list must be terminated with a newline or semicolon.
          This is known as a group command.  The return status is the exit status of list.  Note that unlike the  metachar‐
          acters  (  and  ), { and } are reserved words and must occur where a reserved word is permitted to be recognized.
          Since they do not cause a word break, they must be separated from list by whitespace or another shell metacharac‐
          ter.

Best Answer

Related Solutions

Bash – Rule for Invoking Subshell

Bash Shell – How to Make Explicit Operator Precedence Without Creating a Subshell

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