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Why doesn't foo=bar echo "$foo" print bar?

This is subtle, and has to do with the exact order in which the BashParser performs each step.

Many people, when they first learn about var=value command and how it temporarily sets a variable for the duration of a command, eventually work up an example like this one and become confused why it doesn't do what they expect.

As an illustration:

$ unset foo
$ foo=bar echo "$foo"

$ echo "$foo"

$ foo=bar; echo "$foo"

The reason the first one prints a blank line is because of the order of these steps:

This version works as we expect:

$ unset foo
$ foo=bar bash -c 'echo "$foo"'

In this case, the following steps are performed:

It's not entirely clear, in all cases, why people ask us this question. Mostly they seem to be curious about the behavior, rather than trying to solve a specific problem; so I won't try to give any examples of "the right way to do things like this", since there's no real problem to solve.

There are some special cases in Bash where understanding this can be useful. Take the following examples:

  Array1=( "Var 1" "Var 2" "Var 3" "Var 4" )

  # join each array element with a ";"
  # Traditional solution: set IFS, then unset it afterward
  unset IFS

  # Alternative solution: temporarily set IFS for the duration of eval
  IFS=";" eval 'JoinedVariable="${Array1[*]}"'

Here, the eval alternative is simpler and more elegant. Appropriate care must be taken to ensure safety when using eval.

  # split the string at ";"
  # Traditional solution, using read with temporary IFS value
  IFS=";" read -ra Array2 <<< "$JoinedVariable"

  # Alternative, using eval with temporary IFS value and set -f/+f
  set -f
  IFS=";" eval 'Array2=(${JoinedVariable})'
  set +f

We need set -f to prevent glob expansion of the fields of the JoinedVariable. set +f restores globbing after we're done. In this case, the eval alternative didn't really help us any; in fact, it's quite a bit worse, as we must assume that the script wants globbing enabled.

2012-07-01 04:05