Linux BASH：只有当函数被管道传输时，全局变量才能在函数中更新（简单示例）
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BASH: Global variables aren't updateable in a function only when that function is piped (simple example)
提问by David Parks
This smells buggy, but probably, someone can explain it:
这闻起来有问题，但可能有人可以解释一下：
The following script doesn't work, the output is below:
以下脚本不起作用，输出如下：
#!/bin/bash
GLOBAL_VAR="OLD"
myfunc() {
echo "func before set> $GLOBAL_VAR"
GLOBAL_VAR="NEW"
echo "func after set> $GLOBAL_VAR"
}
myfunc | cat
echo "final value> $GLOBAL_VAR"
Output:
输出：
func before set> OLD
func after set> NEW
final value> OLD
Now, just take off the | catand it works!
现在，只需取下| cat它就可以了！
#!/bin/bash
GLOBAL_VAR="OLD"
myfunc() {
echo "func before set> $GLOBAL_VAR"
GLOBAL_VAR="NEW"
echo "func after set> $GLOBAL_VAR"
}
myfunc
echo "final value> $GLOBAL_VAR"
Output:
输出：
func before set> OLD
func after set> NEW
final value> NEW
采纳答案by Rajish
A pipe creates a subshell. It's said in the bash manualthat subshells cannot modify the environment of their parents. See these links:
管道创建子壳。它在说bash的手册是子shell不能修改其父母的环境。请参阅这些链接：
http://www.gnu.org/software/bash/manual/bashref.html#Pipelines
http://www.gnu.org/software/bash/manual/bashref.html#Pipelines
http://wiki.bash-hackers.org/scripting/processtree#actions_that_create_a_subshell
http://wiki.bash-hackers.org/scripting/processtree#actions_that_create_a_subshell
