8

My POSIX is_integer () function looks like this for a long time:

#!/bin/sh
is_integer ()
{
    [ "$1" -eq "$1" ] 2> /dev/null
}

However, today, I found it broken. If there are some spaces around the number, it surprisingly also evaluates to true, and I have no idea how to fix that.


Example of correct (expected) behavior:

is_integer 123 evaluates to true.

Example of incorrect (unexpected) behavior:

is_integer ' 123' also evaluates to true, however it obviously contains a leading space, thus the function is expected to evaluate to false in such cases.


POSIX-compliant suggestions only, please. Thank you.

  • 1
    Note that even in the absence of spaces, you couldn't use that to detect integers in ksh, since extends the interpretation of the arguments of -eq quite a lot wider than to just integers. Stuff like abc (the value on variable abc), 12.345 (floating point), 1+1 (arithmetic expression) get accepted. – ilkkachu Jul 12 at 10:13
11
#!/bin/sh
is_integer ()
{
    case "${1#[+-]}" in
        (*[!0123456789]*) return 1 ;;
        ('')              return 1 ;;
        (*)               return 0 ;;
    esac
}

Uses only POSIX builtins. It is not clear from the spec if +1 is supposed to be an integer, if not then remove the + from the case line.

It works as follows. the ${1#[+-]} removes the optional leading sign. If you are left with something containing a non digit then it is not an integer, likewise if you are left with nothing. If it is not not an integer then it is an integer.

Edit: change ^ to ! to negate the character class - thanks @LinuxSecurityFreak

| improve this answer | |
  • I don't get why ^ was changed to !. Isn't ^ used to negate the class? See RE bracket expressions – Quasímodo Jul 12 at 11:34
  • True! Just for the record, here the exception is documented in Shell Command Language, Pattern Matching Notation. – Quasímodo Jul 12 at 11:43
  • Your tests should have some multi-digit strings in them. You might want to extend the patterns to reject leading zeros (accept '0', reject '0'*). – icarus Jul 12 at 21:57
  • @icarus. You can replace [!0123456789] with a POSIX class, i.e. [![:digit:]] – fpmurphy Jul 12 at 22:43
  • 1
    @fpmurphy That is true that you can use [:digit:] but I would rather not have 123๔ as an integer because the last character is a digit 4 in Thai. In the original version of this post I use [^0-9] but changed it to explicitly list the characters I wanted to use in the definition of an integer. – icarus Jul 12 at 22:54
1

Not the most efficient (due to the external command), but quite simple:

is_integer () {
  expr "X$1" : "X-\{0,1\}[0-9][0-9]*$" > /dev/null
}

At least in the implementation I am testing, an initial argument - is treated not as part of a matching operation, but apparently as part of an invalid arithmetic expression; the X ensures expr parses its arguments as a valid match operation.

| improve this answer | |

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