I have two shell script like as follows:
a.sh
tes=2
testfunction(){
tes=3
echo 5
}
testfunction
echo $tes
b.sh
tes=2
testfunction(){
tes=3
echo 5
}
val=$(testfunction)
echo $tes
echo $val
In first script tes value is '3' as expected but in second it's 2?
Why is it behaving like this?
Is $(funcall) creating a new sub shell and executing the function? If yes, how can address this?
$() and `` create new shell and return output as a result.
Use 2 variables:
tes=2
testfunction() {
tes=3
tes_str="string result"
}
testfunction
echo $tes
echo $tes_str
output
3
string result
Your current solution creates a subshell which will have its own variable that will be destroyed when it is terminated.
One way to counter this is to pass tes as a parameter, and then return* it using echo.
tes=2
testfunction(){
echo $1
}
val=$(testfunction $tes)
echo $tes
echo $val
You can also use the return command although i would advise against this as it is supposed to be used to for return codes, and as such only ranges from 0-255.Anything outside of that range will become 0
To return a string do the same thing
tes="i am a string"
testfunction(){
echo "$1 from in the function"
}
val=$(testfunction "$tes")
echo $tes
echo $val
i am a string
i am a string from in the function
*Doesnt really return it, it just sends it to STDOUT in the subshell which is then assigned to val
$(funcall)creates a sub-shell. See stackoverflow.com/questions/23951346/…