return value from subshell and output to local variables

Question

I've found the strange behaviour for me, which I can't explain. The following code is work OK:

function prepare-archive {
blah-blah-blah...
_SPEC_FILE=$(check-spec-file "$_GIT_DIR/packaging/")
exit $?
blah-blah-blah...
}

means I get value which I expect:

bash -x ./this-script.sh:
++ exit 1
+ _SPEC_FILE='/home/likern/Print/Oleg/print-service/packaging/print-service.spec
/home/likern/Print/Oleg/print-service/packaging/print-service2.spec'
+ exit 1

As soon as I add local definition to variable:

local _SPEC_FILE=$(check-spec-file "$_GIT_DIR/packaging/")

I get following:

bash -x ./this-script.sh:
++ exit 1
+ local '_SPEC_FILE=/home/likern/Print/Oleg/print-service/packaging/print-service.spec
/home/likern/Print/Oleg/print-service/packaging/print-service2.spec'
+ exit 0
$:~/MyScripts$ echo $?
0

Question: Why? What has happened? Can I catch output from subshell to local variable and check subshell's return value reliably?

P.S.: prepare-archive is called in the main shell script. The first exit is the exit from check-spec-file function, the second from prepare-archive function - this function itself is executed from main shell script. I return value from check-spec-file by exit 1, then pass this value to exit $?. Thus I expect they should be the same.

Answer 1

From the bash manual, Shell Builtin Commands section:

local:
    [...]The return status is zero unless local is used outside a function, an invalid name is supplied, or name is a readonly variable. 

Hope this helps =)

Answer 2

As I use bash subshell parenthesis to group many echo commands I hit this strange problem. In my case all I needed was to pass one value back to the calling shell so I just used the exit command

RET=0
echo RET: $RET
(echo hello
echo there
RET=123
echo RET: $RET
exit $RET)
RET=$?
echo RET: $RET

gives the following output

RET: 0
hello
there
RET: 123
RET: 123

without the exit command you will get this which is confusing:

RET: 0
hello
there
RET: 123
RET: 0

Answer 3

To capture subshell's exit status, declare the variable as local before the assignment, for example, the following script

#!/bin/sh

local_test()
{
    local local_var
    local_var=$(echo "hello from subshell"; exit 1)
    echo "subshell exited with $?"
    echo "local_var=$local_var"
}

echo "before invocation local_var=$local_var in global scope"
local_test
echo "after invocation local_var=$local_var in global scope"

produces the following output

before invocation local_var= in global scope
subshell exited with 1
local_var=hello from subshell
after invocation local_var= in global scope