I have:
func() {
echo a "b c"
}
set $(func)
echo 1: $1
echo 2: $2
echo 3: $3
I want to get two arguments: "a" and "b c". How should func() echo to achieve that?
Tried as above. Getting
1: a
2: b
3: c
I want
1: a
2: b c
3:
4 Answers
How should func() echo to achieve that?
It is not possible to do that. No matter the content of func(), it's impossible with set $(func) to achieve that.
Prefer:
func() {
array=(a "b c")
}
func
set -- "${array[@]}"
Do not use, because eval is evil:
func() {
printf "%q " a "b c"
}
eval "set -- $(func)"
3 Comments
Wolfgang Rohdewald
IMHO array is local to func() and not available after func() finishes.
KamilCuk
I disagree. There is no
local in the function. array is not local and the code works as presented completely fine. In Bash, variables have global (process) scope, unless declared with local or declare in a function.Wolfgang Rohdewald
This surprises me, but you are right.
Bash's mapfile can be used to read a stream of null-delimited values output from a function into an array.
Applied to question:
#!/usr/bin/env bash
func() {
printf '%s\0' a "b c"
}
mapfile -d '' arr < <(func)
echo 1: "${arr[0]}"
echo 2: "${arr[1]}"
echo 3: "${arr[2]}"
Comments
I do not like the array variant, it looks too complicated. But I was able to restructure my code, avoiding this problem.
Comments
2 mistakes:
You forgot to escape the quotes, so your output is:
$ echo a "b c"
a b c
What you want is:
$ echo a \"b c\"
a "b c"
And you have to evaluate the response:
eval set $(func)
That's all
2 Comments
Wolfgang Rohdewald
Does not work for me. The result are still three arguments.
func() { echo a "b c" } eval set $(func) echo 1: $1 echo 2: $2 echo 3: $3 
dekerser
Your comment seems to suggest that you forgot the back slashes in the function. They are essential. So
func() { echo a \"b c\" }
funcisa b c, so how would you know which letters are to be grouped together?