Why does FOO=bar; export the variable into my environment

Question

If I run

FOO=bar docker run -it -e FOO=$FOO debian env

That environment variable is not set in the command output for the env command.

PATH=/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin
HOSTNAME=03f3b59c0aab
TERM=xterm
FOO=
HOME=/root

But if I run

FOO=bar; docker run -i -t --rm -e FOO=$FOO debian:stable-slim env
PATH=/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin
HOSTNAME=672bfdcde93c
TERM=xterm
FOO=bar
HOME=/root

Then the variable is available from the container and also exported into my current shell environment.

echo $FOO
bar

I expect this behavior with export FOO=bar but why does that happen with ; too?

Answer 1

No, FOO=bar; does not export the variable into my environment

A var is set in the (present) environment only if it was previously exported:

$ export foo
$ foo=bar
$ env | grep foo
foo=bar

A variable is set in the environment of a command when it is placed before the command. Like foo=bar command. And it only exists while the command runs.

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

The var is not set for the command line (in the present shell):

$ foo=bar bash -c echo\ $foo

Above, the value of $foo is replaced with nothing by the present running shell, thus: no output.

Your command:

$ FOO=bar docker run -it -e FOO=$FOO debian env

is converted to the actual string:

$ FOO=bar docker run -it -e FOO= debian env

by the present running shell.

If, instead, you set the variable (in the present running shell) with foo=bar before running the command, the line will be converted to:

$ FOO=bar; docker run -it -e FOO=bar debian env

A variable set to the environment of a command is erased when the command returns:

$ foo=bar bash -c 'echo'; echo "foo was erased: \"$foo\""

Except when the command is a builtin in some conditions/shells:

$ ksh -c 'foo=bar typeset baz=quuz; echo $foo'
bar

Answer 2

There are a number of variations to consider:

1. Just doing FOO=bar creates a variable named FOO with value bar, but that variable isn't passed along to new processes:

   $ echo $FOO
   $ FOO=bar
   $ echo $FOO
   bar
   $ bash        # Start a new bash process
   $ echo $FOO
                 # Variable is not set in the new process
   $ exit        # Exit new bash process
   

2. Running FOO=bar <command> will run the given command with the variable set (but doesn't affect the original shell's environment):

   $ echo $foo
   $ FOO=baz bash   # start a new bash process
   $ echo $FOO
   baz
   $ exit           # exit the new bash process
   exit
   $ echo $FOO      
                    # No FOO in the original bash process
   $
   

3. Doing FOO=foo; <command> is equivalent to (1); putting a semicolon between two commands is equivalent to running those two commands on two separate lines:

   $ FOO=foo; echo $FOO
   foo
   $ bash
   $ echo $FOO
   
   $ exit
   exit
   $ echo $FOO
   foo
   $
   

4. Using export will pass a variable in the shell's environment to newly-created processes:

   $ export FOO=bar
   $ echo $FOO    # Value set in original shell
   bar
   $ bash         # Start another shell
   $ echo $FOO
   bar            # Value was passed along to new process
   $ exit
   exit
   

Answer 3

FOO=bar docker run -it -e FOO=$FOO debian env

Here the $FOO from FOO=$FOO will be expanded before the FOO=bar assignment happens.

You can check that with a more straight-forward example:

FOO=first
FOO=second echo FOO=$FOO
=> FOO=first

FOO=third; echo FOO=$FOO
=> FOO=third

The FOO=bar cmd form will really set FOO=bar in the environment of cmd, but a command like docker does not automatically export its own environment into the container, but the environment vars have to be added explicitly with the -e switch.

Again, a more straightforward demo would be:

FOO=before
FOO=after env - FOO=$FOO printenv FOO
=> before