Why bash -c 'var=5 printf "$var"' does not print 5? [duplicate]

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I tried these variations, but none of them do nothing, i.e, do not print 5

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linux@linux:~$ bash -c 'var=5 printf "$var"'
linux@linux:~$ bash -c "export var=5 echo $var"
linux@linux:~$ bash -c "export var=5 && echo $var"

linux@linux:~$ bash -c "var=5; echo $var"
linux@linux:~$ var=5 sudo printf "$var"
linux@linux:~$ var=5 sudo -E printf "$var"

linux@linux:~$ bash --version
GNU bash, version 4.4.19(1)-release (x86_64-pc-linux-gnu)
Copyright (C) 2016 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.
linuxbashshellcommand-lineenvironment-variables
Some of them came close to working, except for the usage of the wrong kind of quotes. bash -c 'var=5; echo "$var"' works fine, f/e. However, when you pass the code to bash -c in double quotes, the $var gets replaced with the value of var in the parent shell, before the copy of bash passed the -c argument is even started, and thus before the assignment takes place. - Charles Duffy
...I've added a second duplicate to the list covering that case. - Charles Duffy
export var=5 echo $var exports a variable named echo. The related expression var=5 echo $var calls echo with either one argument (the value of $var in the caller) or no arguments (if $var is empty or unset in the caller) and the environment variable set to 5. But echo ignores its environment, so that doesn't change its behavior. - William Pursell
var=5 is stored into the environment of the called program (here echo or printf). But $val is evaluated before the program is called. val=5 sh -c 'echo $val' or val=5 sh -c "echo \$val" will work, because the called program sh will expand $val. - Wiimm