How do I check if user input is an integer in Ruby?

Question

I am trying to loop until user inputs an integer. When user inputs a letter, the following code should print "Think of a number":

print "Think of a number "

while user_input = gets.to_i
  if user_input.is_a? Integer
    puts "your number is #{user_input}"
    break
  else
    print "Think of a number "
  end 
end

I succeeded with my code when user inputs an integer. However when user inputs a string, the to_i method returns 0, and does not execute the else statement because it is a number.

Maybe this from @sawa: stackoverflow.com/a/14900559/5239030 helps? — iGian
– iGian, Commented Feb 19, 2019 at 8:26

Aleksei Matiushkin · Accepted Answer · 2019-02-19 08:19:15Z

4

The main issue with your code is String#to_i method is omnivorous.

"0".to_i #⇒ 0
"0.1".to_i #⇒ 0
"foo".to_i #⇒ 0

That said, user_input in your code is always integer.

What you probably want is to accept digits only (and maybe a leading minus for negatives.) The only concise way to accept a subset of characters is a regular expression.

# chomp to strip out trailing carriage return
user_input = gets.chomp 

if user_input =~ /\A-?\d+\z/
  ...

The regular expression above means nothing save for digits with optional leading minus.

Or, even better (credits to @Stefan)

if gets =~ /\A-?\d+\Z/

edited Feb 19, 2019 at 8:19

answered Feb 19, 2019 at 6:29

Aleksei Matiushkin

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4 Comments

sawa Over a year ago

Don't always use the positive form for matching. Use !user_input.match?(/\D/).

sawa Over a year ago

This is old-school way. Nowadays, you can use Integer method.

Aleksei Matiushkin Over a year ago

@sawa yes, Integer(exception: false) but Ruby 2.6 is not yet everywhere. Also, /\D/ would hardly allow negatives.

Stefan Over a year ago

There's also \Z (uppercase) which allows you to omit chomp.

Oli Crt · Accepted Answer · 2019-02-19 09:39:35Z

2

If you only want to accept postive digits, you can use a range:

user_input = gets.chomp 
if ('0'..'9').cover? user_input

answered Feb 19, 2019 at 9:39

Oli Crt

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Comments

rahul mishra · Accepted Answer · 2019-02-19 03:35:31Z

0

let check below one used Integer(gets.chomp) rescue ''

print "Think of a number "

while user_input = Integer(gets.chomp) rescue ''
  if user_input.is_a? Integer
    puts "your number is #{user_input}"
    break
  else
    print "Think of a number "
  end
end

answered Feb 19, 2019 at 3:35

rahul mishra

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4 Comments

Aleksei Matiushkin Over a year ago

Integer(get.chomp, exception: false) since Ruby 2.6.

Stefan Over a year ago

You can omit chomp here, but you probably want to set the base to 10.

rahul mishra Over a year ago

@Stefan I am not sure what exactly you mean, because by default the base is 10

Stefan Over a year ago

to_i has a default base of 10 whereas Integer() has a default base of 0, i.e. it attempts to determine the base from the input: Integer('0100') #=> 64

Vaibhav · Accepted Answer · 2019-02-19 05:18:42Z

-1

I came across a similar problem. I ended up doing this:

if user_input.strip == user_input.to_i.to_s
  # More code here!
end

Testing for float would be:

if user_input.strip == user_input.to_f.to_s
  # More code here!
end

Solved my issue. See if it helps.

answered Feb 19, 2019 at 5:18

Vaibhav

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4 Comments

Aleksei Matiushkin Over a year ago

The latter would eliminate integers though.

Vaibhav Over a year ago

Exactly. It checks only for floats. If one needs both, they can be ored.

Aleksei Matiushkin Over a year ago

Technically 1 is somewhat float since 1 == 1.0 #⇒ true.

Aleksei Matiushkin Over a year ago

Also it won’t work on absolutely legit 01 for integers and 1.00 for floats.

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