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.

Dupe of https://stackoverflow.com/questions/1235863/test-if-a-string-is-basically-an-integer-in-quotes-using-ruby? — orde, Feb 19 '19 at 03:19
Maybe this from @sawa: https://stackoverflow.com/a/14900559/5239030 helps? — iGian, Feb 19 '19 at 08:26

Aleksei Matiushkin · Accepted Answer · 2019-02-19T08:19:15.417

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 '19 at 08:19

answered Feb 19 '19 at 06:29

Aleksei Matiushkin

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Don't always use the positive form for matching. Use `!user_input.match?(/\D/)`. – sawa Feb 19 '19 at 07:20
1

This is old-school way. Nowadays, you can use `Integer` method. – sawa Feb 19 '19 at 07:21
1

@sawa yes, `Integer(exception: false)` but Ruby 2.6 is not yet everywhere. Also, `/\D/` would hardly allow negatives. – Aleksei Matiushkin Feb 19 '19 at 07:31
2

There's also `\Z` (uppercase) which allows you to omit `chomp`. – Stefan Feb 19 '19 at 08:14

score 2 · Answer 2 · answered Feb 19 '19 at 09:39

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 '19 at 09:39

Oli Crt

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score 0 · Answer 3 · answered Feb 19 '19 at 03:35

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 '19 at 03:35

rahul mishra

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2

`Integer(get.chomp, exception: false)` since Ruby 2.6. – Aleksei Matiushkin Feb 19 '19 at 05:52
2

You can omit `chomp` here, but you probably want to set the base to `10`. – Stefan Feb 19 '19 at 07:49
@Stefan I am not sure what exactly you mean, because by default the base is 10 – rahul mishra Feb 19 '19 at 07:56
2

`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` – Stefan Feb 19 '19 at 07:59

score -1 · Answer 4 · answered Feb 19 '19 at 05:18

-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 '19 at 05:18

Vaibhav

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The latter would eliminate integers though. – Aleksei Matiushkin Feb 19 '19 at 06:23
Exactly. It checks only for floats. If one needs both, they can be `or`ed. – Vaibhav Feb 19 '19 at 06:28
1

Technically `1` is somewhat float since `1 == 1.0 #⇒ true`. – Aleksei Matiushkin Feb 19 '19 at 06:30
1

Also it won’t work on absolutely legit `01` for integers and `1.00` for floats. – Aleksei Matiushkin Feb 19 '19 at 06:32

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

4 Answers4