Bash – UNIX export command

bashshellunix

I am trying to understand the use of export command.

I tried using man export, but there is no manual for this command.

Can anyone please help me out understanding the use of export in UNIX?

Best Solution

When you execute a program the child program inherits its environment variables from the parent. For instance if $HOME is set to /root in the parent then the child's $HOME variable is also set to /root.

This only applies to environment variable that are marked for export. If you set a variable at the command-line like

$ FOO="bar"

That variable will not be visible in child processes. Not unless you export it:

$ export FOO

You can combine these two statements into a single one in bash (but not in old-school sh):

$ export FOO="bar"

Here's a quick example showing the difference between exported and non-exported variables. To understand what's happening know that sh -c creates a child shell process which inherits the parent shell's environment.

$ FOO=bar
$ sh -c 'echo $FOO'

$ export FOO
$ sh -c 'echo $FOO'
bar

Note: To get help on shell built-in commands use help export. Shell built-ins are commands that are part of your shell rather than independent executables like /bin/ls.

Related Question