Why do following output `$var` instead of `3`? [closed]

Why do the following examples from Arrow's reply output $var instead of 3?

I single quote or backslash $var as '$var' or \$var, hoping that

• first $var will be passed literally into the execution environment of the executable program /bin/echo, and
• then inside the execution environment of the executable program /bin/echo, the environment variable var=3 applies when parameter expansion happens on $var.

Why doesn't the parameter expansion happen inside the execution environment of the executable program /bin/echo? Note:

• in the original shell, $var is single quoted or backslashed, so should not be expanded in the execution environment of the original shell.

• the original shell removes the single quotes or backslash around $var, and then passes $var and environment variable var=3 into the execution environment for running /bin/echo, so I think that parameter expansion of $var should happen in the execution environment for running /bin/echo.

Note that I use set -x to print out the trace information, but can't figure out what is actually the result after expansion but before execution.

Thanks.

tim$ unset var $ set -x tim$ var=3 /bin/echo '$var' + var=3 + /bin/echo '$var' $var tim$ var=3 /bin/echo \$var + var=3 + /bin/echo '$var' $var $ var=3 exec /bin/echo \$var | cat + cat + var=3 + exec /bin/echo '$var' $var tim$ var=3 exec /bin/echo '$var' | cat + var=3 + exec /bin/echo '$var' + cat $var