check if parameter is integer with numeric expression

Question

The following program checks if the first parameter is a number, or a not a number. The best I could come up with so far is:

#/bin/bash
if (( $# )) && (( $1 != 0 )) ; then
    echo "number"
else
    echo "not number"
fi

This works for numbers like 1, -1, 100, etc, for strings a, abc, a431, or if no parameter is given.
It works with a warning for strings like 1a1, 1.3, -1.2
But FAILS (obviously) if the parameter is 0

How could this be fixed? (using arithmetic expressions)

Related: Bash integer comparison

possible duplicate of BASH: Test whether string is valid as an integer? — tripleee
– tripleee, Commented Sep 4, 2013 at 17:35

chepner · Accepted Answer · 2013-01-25 14:19:38Z

2

You can press the printf built-in command to do the job:

if printf "%d" ${1:-foo} > /dev/null; then
    echo "number"
else
    echo "NOT a number"
fi

printf exits with non-zero status if the conversion fails. The ${1-...} notation provides a known not-number if $1 isn't set.

answered Jan 25, 2013 at 14:19

chepner

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Comments

DreadPirateShawn · Accepted Answer · 2013-01-25 03:49:48Z

1

You could use a regex:

#/bin/bash

if [[ "$1" =~ ^-?[0-9]+$ ]] ; then
    echo "number"
else
    echo "not number"
fi

If you wanted it to handle floating point, you could tweak it easily enough:

#/bin/bash

if [[ "$1" =~ ^-?[0-9]+$ || "$1" =~ ^-?[0-9]*\.[0-9]+ ]] ; then
    echo "number"
else
    echo "not number"
fi

answered Jan 25, 2013 at 3:49

DreadPirateShawn

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1 Comment

user000001 Over a year ago

yes this passes all the test I could think of. Not accepting yet though waiting for better answers (if any).

vpillai · Accepted Answer · 2013-01-25 03:22:47Z

1

I would rather use something like

if [ $1 = "$(echo $1 | awk '{print strtonum($0)}')" ]; then
echo "number"
else
echo "not number"
fi

answered Jan 25, 2013 at 3:22

vpillai

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2 Comments

user000001 Over a year ago

Yes this would work. But I will give it some time in case someone will be able to solve it with the above constrains.

user000001 Over a year ago

Also it throws a warning if no parameter is given and identifies floats as numbers

koola · Accepted Answer · 2013-01-25 05:52:06Z

1

Solution using extended globs

#/bin/bash
shopt -s extglob
if [[ $1 = *[0-9]* && $1 = ?([+-])*([0-9])?(.*([0-9])) ]]; then
    echo "Is a number"
else
    echo "Not a number"
fi

Output

$ ./test -1
Is a number
$ ./test 0
Is a number
$ ./test 1
Is a number
$ ./test a
Not a number
$ ./test -1.2
Is a number
$ ./test 0.0
Is a number
$ ./test 1.2
Is a number
$ ./test a.b
Not a number

answered Jan 25, 2013 at 5:52

koola

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4 Comments

DreadPirateShawn Over a year ago

Plus 1 for prompting more people to Google extended globbing. Minus 1 (IMHO) for the additional complexity, and a regex-style matcher that can't be directly translated if the script gets ported to something other than Bash. Granted, YMMV.

koola Over a year ago

I disagree, this solution is no more complex or harder to port than yours.

DreadPirateShawn Over a year ago

Globbing is Bash-only syntax that doesn't quite match standard regex structure. I didn't say it was vastly more difficult, just that there's a certain utility in KISS when you don't specifically need Bash-specific features, and can use a generalized regex structure instead.

DreadPirateShawn Over a year ago

But not to worry, it's not like an answer is going to be chosen on this question anyway. :-)

William · Accepted Answer · 2013-01-25 17:55:54Z

1

Might be easier to test for what is not a number:

if (( $# )) ; then
  if [[ $1 == *[^0-9]* ]] ; then
    echo "Not a number"
  else
    echo "Is a number"
  fi
fi

Note that the above is actually testing only for positive integers. Add +-. to the character class and it will work, sort of, for positive/negative integers and decimal fractions. I say sort of because it would pass 123.456.789 as a number, but it's not valid. You'd need to add additional logic to weed out misplaced or multiple signs and decimal points (and strings like "+", ".", etc.).

edited Jan 25, 2013 at 17:55

answered Jan 25, 2013 at 14:54

William

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