從subshel​​l返回值和輸出到局部變量

Question

我發現了我的奇怪行爲，我無法解釋。 下面的代碼是工作確定：

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

意味着我得到我期待值：

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 

只要我添加local定義變量：

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

我獲得以下：

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 

問題：爲什麼？發生了什麼？我能否將subshel​​l的輸出捕獲到local變量並可靠地檢查子殼的返回值？

P.S.在主shell腳本中調用：prepare-archive。第一個exit是exit從check-spec-file函數，第二個從prepare-archive函數 - 這個函數本身是從主shell腳本執行的。我將check-spec-file的值返回exit 1，然後將此值傳遞給exit $?。因此我希望他們應該是一樣的。

Answer 1

從bash的手冊，Shell Builtin Commands部分：

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

希望這有助於=）

Answer 2

要捕獲子shell的退出狀態，分配前變量聲明爲局部的，例如，下面的腳本

#!/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" 

產生以下輸出

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