If I do this:


I get no output

but if I do:


then I get


why is that?

marked as duplicate by Gilles bash Dec 11 '16 at 23:00

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In command line you need to separate commands using either ; or &&. Space it is just not a delimiter for CLI commands.

With && the second command will be executed if the first command exits successfully (exit code 0).

With ; the second command will be executed no matter what is the exit status of first command.


# a1;echo hello
bash: a1: command not found
# a1 && echo hello
bash: a1: command not found

In scripts the first command and the second command is separated by a new line \n character, also recognized by bash.

  • 1
    A variable assignment is not a command. The reason for OP's behavior is because the expansion happens before the assignment. – jordanm Dec 11 '16 at 2:12

SUMAN_DEBUG=foo is a local environment variable assignment for a local variable that exists only for the command it prefaces. In your second case, there is no command because the && initiates a list of command where in your case, the first command is null (so the environment variable has no command to affect) and the second is executed, but has no environment variable set for it.

You could add export to the assignment so that the variable would become a global environment variable. But then it would persist after the list of commands is complete.

Another option is to put the entire command list in parenthesis to execute it in a subshell:

(export SUMAN_DEBUG=foo && echo $SUMAN_DEBUG)

In this case, the persistent environment variable would be set in a subshell and that environment would disappear when the subshell completed its work).

See the Bourne-again Shell Manual, lists section and the section on Environment for more.

  • It looks like this also works well without the export: (SUMAN_DEBUG=foo && echo $SUMAN_DEBUG) and keeps it in the scope of the command sequence – groovenectar Mar 27 at 15:30

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