bash shell 脚本和脚本中函数的变量范围
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/21006674/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Variable scope for bash shell scripts and functions in the script
提问by Classified
I'm a little confused with my script regarding functions, variable scope, and possibly subshells. I saw in another postthat pipes spawn a subshell and the parent shell can't access variables from the subshell. Is this the same case with cmds run in backticks too?
我对我的脚本关于函数、变量范围和可能的子外壳有点困惑。我在另一篇文章中看到管道产生了一个子外壳，而父外壳不能从子外壳访问变量。这与 cmds 在反引号中运行的情况相同吗？
To not bore people, I've shortened my 100+ line script but I tried to remember to leave in the important elements (i.e. backticks, pipes etc). Hopefully I didn't leave anything out.
为了不让人厌烦，我缩短了 100 多行脚本，但我尽量记住保留重要元素（即反引号、竖线等）。希望我没有遗漏任何东西。
global1=0
global2=0
start_read=true
function testfunc {
global1=9999
global2=1111
echo "in testfunc"
echo $global1
echo $global2
}
file1=whocares
file2=whocares2
for line in `cat $file1`
do
for i in `grep -P "\w+ stream" $file2 | grep "$line"` # possible but unlikely problem spot
do
end=$(echo $i | cut -d ' ' -f 1-4 | cut -d ',' -f 1) # possible but unlikely spot
duration=`testfunc $end` # more likely problem spot
done
done
echo "global1 = $global1"
echo "global2 = $global2"
So when I run my script, the last line says global1 = 0. However, in my function testfunc, global1 gets set to 9999 and the debug msgs print out that within the function at least, it is 9999.
所以当我运行我的脚本时，最后一行说 global1 = 0。但是，在我的函数 testfunc 中，global1 被设置为 9999 并且调试消息至少在函数内打印出来，它是 9999。
Two questions here:
这里有两个问题：
- Do the backticks spawn a subshell and thus making my script not work?
- How do I work around this issue?
- 反引号是否会产生一个子shell，从而使我的脚本不起作用？
- 我该如何解决这个问题？
Thanks in advance for your help.
在此先感谢您的帮助。
采纳答案by damienfrancois
You can try something like
你可以尝试类似的东西
global1=0
global2=0
start_read=true
function testfunc {
global1=9999
global2=1111
echo "in testfunc"
echo $global1
echo $global2
duration=something
}
file1=whocares
file2=whocares2
for line in `cat $file1`
do
for i in `grep -P "\w+ stream" $file2 | grep "$line"` # possible but unlikely problem spot
do
end=$(echo $i | cut -d ' ' -f 1-4 | cut -d ',' -f 1) # possible but unlikely spot
testfunc $end # more likely problem spot
done
done
echo "global1 = $global1"
echo "global2 = $global2"
回答by anubhava
Do the backticks spawn a subshell and thus making my script not work?:
Do the backticks spawn a subshell and thus making my script not work?：
Yes they do and any changes made in variable in a subshell are not visible in parent shell.
是的，在子 shell 中对变量所做的任何更改在父 shell 中都不可见。
How do I work around this issue?
How do I work around this issue?
You can probably try this loop that avoid spawning a subshell:
你可以试试这个避免产生子shell的循环：
while read line
do
while read i
do
end=$(echo $i | cut -d ' ' -f 1-4 | cut -d ',' -f 1)
duration=$(testfunc "$end")
done < <(grep -P "\w+ stream" "$file2" | grep "$line")
done < "$file1"
PS: But testfuncwill still be called in sub process.
PS：但testfunc仍会在子进程中调用。
