SC2097

This assignment is only seen by the forked process.

Problematic code:

name=World cmd -m "Hello $name"

Correct code:

name=World
cmd -m "Hello $name"

If the original goal was to limit the scope of the variable, this can also be done in a subshell:

(
   name=World
   cmd -m "Hello $name"
) # 'name' does not leave this subshell

Rationale:

In name=World cmd "$name", name=World is passed in as part of the environment to cmd (i.e., in the envp parameter to execve(2)). This means that cmd and its children will see the parameter, but no other processes will.

However, "$name" is not expanded by cmd. "$name" is expanded by the shell before cmd is ever executed, and thus it will not use the new value.

The solution is to set the variable first, then use it as a parameter. If limited scope is desired, a ( subshell ) can be used.

Exceptions

In the strange and fabricated scenarios where the script and a program uses a variable name for two different purposes, you can ignore this message. This is hard to conceive, since scripts should use lowercase variable names specifically to avoid collisions with the environment.

Related resources:

• BashFaq: Why doesn't foo=bar echo "$foo" print bar?
• StackOverflow: Prefixing variable assignment doesn't work with echo