bash - How to declare a local integer?

Question

In Bash, how do I declare a local integer variable, i.e. something like:

func() {
  local ((number = 0)) # I know this does not work
  local declare -i number=0 # this doesn't work either

  # other statements, possibly modifying number
}

Somewhere I saw local -i number=0 being used, but this doesn't look very portable.

Answer 1

declare inside a function automatically makes the variable local. So this works:

func() {
    declare -i number=0

    number=20
    echo "In ${FUNCNAME[0]}, \$number has the value $number"
}

number=10
echo "Before the function, \$number has the value $number"
func
echo "After the function, \$number has the value $number"

And the output is:

Before the function, $number has the value 10
In func, $number has the value 20
After the function, $number has the value 10