command substitution within c style for loop in Zsh

Question

I have a function printing a number

myFunc() {
    echo 11
}

Now I want to use the number in expression in c style for-loop:

for (( i=1; i<=$(myFunc); i++ )); do
    echo "$i"
done

Zsh gave me

test.sh:5: parse error near `i++ '
test.sh:6: parse error near `i<=$(myFunc); i++ ))...'

while Bash prints 1 through 11 as expected. What happened to Zsh? I expect it to be compatible with Bash on this feature.

In case it matters, my environment:

• OS X 10.11.5
• zsh 5.0.8 (x86_64-apple-darwin15.0)
• GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin15)

update

This may be a bug, and may have been fixed in latest version. I will delete this question as soon as I verify this.

Answer 1

It works in my zsh:

$ zsh --version
zsh 5.7.1 (x86_64-pc-linux-gnu)
$ myFunc() {
    echo 11
}
$ for (( i=1; i<=$(myFunc); i++ )); do
    echo "$i"
done
1
2
3
4
5
6
7
8
9
10
11
$ 

It might just have been a bug.