Assign command's return code to variable in Bash

Question

I have this simple (even silly :) problem: Inside a bash script, I'm trying to store a command's return code to a variable using S?, so as to use it later in the script, but S? is stored literally as S? .
I run this test bash script:

#!/bin/bash
echo "trying to store this command's return code. it should be 0"
rtrn_code=S?
echo $rtrn_code

but instead of getting 0 (success return code) I get S? :

trying to store this command return code. it should be 0
S?

What am I doing wrong? Any ideas?

Thank you in advance.

Gardro, in bash all variable references begin with the $ symbol. The ? is simply a special variable in bash holding the return value from the previous command. So in order to (de)reference the value it holds you must precede it with the $ sign -- $?. The downvotes (not mine) are simply a reflection that before asking here, you are supposed to use diligence to find the answer yourself. This being a fundamental bash-101 issue -- you are going to receive some... — David C. Rankin
– David C. Rankin, Commented Apr 30, 2015 at 9:27

Sachin Joseph · Accepted Answer · 2016-04-12 09:23:03Z

5

It's $? and not S?.

rtrn_code=$?

edited Apr 12, 2016 at 9:23

answered Apr 30, 2015 at 8:43

Sachin Joseph

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iqstatic · Accepted Answer · 2015-04-30 09:48:56Z

2

The return value of the last executed command can be accessed using $?. Replace S? to $? in your script.

edited Apr 30, 2015 at 9:48

answered Apr 30, 2015 at 8:45

iqstatic

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