Using "declare" in Zsh

Question

I am following this thread: https://stackoverflow.com/a/19742842/5057251 for typeset (or declare) in ZSH, not BASH.

#Declare (or typeset) an array of integers
#declare -ai int_array
typeset -ai int_array
int_array=(1 2 3)
echo "${int_array[@]}"

Then

# Attempt to change 1st element to string. (expect to fail)
int_array[1]="Should fail" || echo "error: ${LINENO}"
echo "${int_array[@]}"

Bash finds the error, gracefully reports error and lineno, prints:

1 2 3

But Zsh accepts, prints:

Should fail 2 3

Not sure why different.

Answer 1

There are two problems here:

1. In bash, and zsh, assigning a string to an integer variable causes that string to be evaluated as an arithmetic expression. Thus, this is not an error:

   $ typeset -i foo
   $ foo="bar"
   

   If bar was a variable previously set to an arithmetic expression, then bar's expansion would be evaluated as such:

   $ bar=10+2
   $ typeset -i foo
   $ foo="bar"
   $ echo "$foo"
   12
   

   The error in your assignment, of course, is that there's no way to expand Should fail like that. If it were, say, Should - fail (an arithmetic expression subtracting the value of the two variables Should and fail, for example, it would still work:

   $ foo="Should - fail"
   $ echo "$foo"
   0
   

2. The second problem is that nothing in the zsh docs indicate that -i may be set for an entire array, and so the -a in -ai is ignored:

   bash-5.0$ typeset -ai foo
   bash-5.0$ declare -p foo
   declare -ai foo=([0]="0")  # the previous value was retained in the array
   

   vs zsh:

   % typeset -ai foo
   % foo[1]=10
   % foo[2]=20
   % declare -p foo
   typeset -i foo=20  # treated as a normal variable, not array
   

   What you're seeing is essentially int_array being redeclared as an array (without any qualifiers) when you do int_array=(1 2 3):

   % foo=(1 2 3)
   % declare -p foo
   typeset -a foo=( 1 2 3 )