I'm working with this:
GNU bash, version 4.1.2(1)-release (x86_64-redhat-linux-gnu)
I have a script like below:
#!/bin/bash
e=2
function test1() {
e=4
echo "hello"
}
test1
echo "$e"
Which returns:
hello
4
But if I assign the result of the function to a variable, the global variable e
is not modified:
#!/bin/bash
e=2
function test1() {
e=4
echo "hello"
}
ret=$(test1)
echo "$ret"
echo "$e"
Returns:
hello
2
I've heard of the use of eval in this case, so I did this in test1
:
eval 'e=4'
But the same result.
Could you explain me why it is not modified? How could I save the echo of the test1
function in ret
and modify the global variable too?
When you use a command substitution (ie the $(...)
construct), you are creating a subshell. Subshells inherit variables from their parent shells, but this only works one way - a subshell cannot modify the environment of its parent shell. Your variable e
is set within a subshell, but not the parent shell. There are two ways to pass values from a subshell to its parent. First, you can output something to stdout, then capture it with a command substitution:
myfunc() {
echo "Hello"
}
var="$(myfunc)"
echo "$var"
Gives:
Hello
For a numerical value from 0-255, you can use return
to pass the number as the exit status:
mysecondfunc() {
echo "Hello"
return 4
}
var="$(mysecondfunc)"
num_var=$?
echo "$var - num is $num_var"
Gives:
Hello - num is 4