How to Make `local` Capture the Exit Code in Bash

bashexit-statusvariable

In my project I have the following snippet:

local output="$(bash "${1##*/}")"
echo "$?"

This always prints zero due to local, however, removing local causes the $? variable to behave correctly: which is to assume the exit code from the subshell.

My question is: how I can keep this variable local whilst also capturing the exit value?

Best Answer

#!/bin/bash
thing() {
   local foo=$(asjkdh) ret="$?"
   echo "$ret"
}

This will echo 127, the correct error code for "command not found".

You can use local to define more than one variable. So I just also create the local variable RET to capture the exit code of the subshell before local succeeds and sets $? to zero.

Related Solutions

Bash – How will bash evaluate following code

That's:

if
  first list of commands
then
  second list of commands
else
  third list of commands
fi

That's to run the second list of commands if the first list of commands returns with a true/success (zero) exit status, that is if the last run command in there returns with a zero exit status.

In:

var=-2 && ((var += 2))

It's cmd1 && cmd2 where cmd2 is only run if cmd1 is successful.

var=-2

Will typically be successful as long as $var has not been made read-only, so the ((var += 2)) command will be run:

((arithmetic expression))

Returns success/true as long as the expression is correctly evaluated (no syntax error) and the result of the expression is non-zero.

((123)), ((1 + 1)), ((1 == 1)) return true
((0)), ((-2 + 2)), ((2 == -2)) return false.
_{((4294967296 * 4294967296))}_{return false in most shells because of 64 bit integer wrapping}

var += 2 as an arithmetic expression, performs the assignment and resolves to the value being assigned, here 0, hence the false exit status.

You can see the value upon which is based the exit status, by using the $((...)) arithmetic expansion syntax:

$ echo "$((1 + 1)) $((2 == 2)) $((2 == -2)) $((var = -2)) $((var += 2))"
2 1 0 -2 0

Or assigning it to a variable:

$ var=-2; ((result = (var += 2)))
$ echo "$? $result $var"
1 0 0

$? contains the exit status of the previous command. As far as if/then/else/fi is concerned, 0 means true, anything else means false.

The confusion here comes from the fact that for arithmetic expressions, it's the other way round: 0 means false and anything else means true (for instance, 2 == 2 is 1 while 2 < 1 is 0).

To avoid worrying about the difference, just forget about $? and its possible values. Just think in terms of boolean true/false, sucess/failure.

 grep -q foo file

Returns true if foo is found in file.

 [ "$a" = "$b" ]

Returns true if $a contains the same thing as $b.

 ((6 * 3 - 12))
 ((4 == 1))

Return true if the result of the arithmetic expression is a non-zero number.

It doesn't matter whether those true/false are expressed in terms of 0 or 1 of the exit status of those grep/[ commands or ((...)) construct.

Why Isn’t [ -n ] False Like [ -n “” ] in Bash?

[x] is equivalent to [ -nx] even if x starts with - provided there is no operand.

$ [ -o ] ; echo $?
0
$ [ -eq ] ; echo $?
0
$ [ -n -o ] ; echo $?
0
$ [ -n -eq ] ; echo $?
0

Best Answer

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Bash – How will bash evaluate following code

Why Isn’t [ -n ] False Like [ -n “” ] in Bash?

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