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"
bar

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

This version works as we expect:

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

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 example:

  arr=('Var 1' 'Var 2' 'Var 3' 'Var 4')

  # join each array element with a ";"
  # Traditional solution: set IFS, then unset it afterward
  IFS=\;
  joinedVariable="${arr[*]}"
  unset -v IFS

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

Here, the eval alternative is simpler and more elegant. Appropriate care must be taken to ensure safety when using eval. The command prefix is required for all shells other than Bash plus Bash POSIX mode (see http://wiki.bash-hackers.org/commands/builtin/eval#using_the_environment). This won't work in recent versions of zsh due to an apparent regression (documented behavior broken for setopt POSIX_BUILTINS), or busybox due to a bug in that environment assignments fail to propagate in this case. (feel free to file bugs if anybody cares enough.)