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I have this code where I am entering input for sides of a triangle. Depending on the values, it will print it the triangle is equilateral, isoceles, or scalene. It's executing for number values, but how do I specify that the input should only be integers? For example, if I type in "w" , it should say invalid or error, but in this case, it executes. How do I solve this?

Basically, I am looking for a way to write that if a string were to be inputted, it should show up as an error (then I would write a print statement saying it is invalid). So could I put that into another if statement? (before the ones mentioned below)

Example Code:

puts "Enter the triangle length"
  x = gets.chomp

  puts "Enter the triangle width"
  y = gets.chomp


  puts "Enter the triangle height"
  z = gets.chomp


  if x == y and y == z
      puts "This triangle is equilateral"
      else if
       x==y or y == z or x==z
           puts "This triangle is isoceles"
        else if
          x!=y and y!=z and x!=z
             puts "this triangle is scalene"
         end
      end
  end
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5
  • Where's the Java part? Commented Jun 28, 2016 at 4:36
  • Are you sure you didn't mean elsif instead of else if? Commented Jun 28, 2016 at 4:43
  • What is the difference between elseif and else if? Commented Jun 28, 2016 at 4:44
  • else if is two separate statements. You're putting an if expression inside the else block, i.e. if … else (if … end) end. Note the two ends. That's entirely different from if … else … end and behaves very differently. Commented Jun 28, 2016 at 5:27
  • If you fix your newlines and indentation, the difference becomes very clear: gist.github.com/jrunning/df00de328d53a2a5b67db96616a8f870 In this case the behavior is actually the same, but that's only by accident. Commented Jun 28, 2016 at 5:32

6 Answers 6

Reset to default
2

If you are dealing with integers, you can check this with ruby.

Note, this is not as robust as regex, but it covers most cases.

if (input != '0') && (input.to_i.to_s != input.strip)
  # input is not a number
else
  # input is a number
end

strip is there if you want to accept input with leading or trailing whitespace, otherwise you can leave it off.

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3 Comments

So this if statement is basically saying that as long as the input is still a number and if a string can be converted to an integer, it should accept and execute?
I advise you to open irb and play with to_i, it might surprise you, especially when you call to_i on string. In short, calling to_i on string returns 0. Does my if condition make sense now?
I originally wrote under each variable (X = x.to_i etc) but however, I was noticing that even if I entered a letter, it somehow entered that if loop because it was being recognized as an integer. What would regex do?
1

While all the other answers are probably more or less robust, I would go with another one. Since you have a triangle sides lengths, they are to be greater than zero, right? Then one might use the side effect of String#to_i method: for everything that is not converting to integer it returns zero. Therefore:

x = gets.chomp.to_i
y = gets.chomp.to_i
z = gets.chomp.to_i

raise "Wrong input" unless x > 0 && y > 0 && z > 0

# ...
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2 Comments

I tried the raise command and its working for me. However, would I be able to put in a print statement? I.E puts "Invalid input" (without an error coming from the compiler?)
Of course, just change raise to puts or whatever.
1

You can do something like this:

x = x.to_i

puts "Please enter an integer" if x == 0

Why?

Because:

"ABC".to_i # returns 0

You may be better off calling strip instead of chomp

gets.strip.to_i

An example:

## ruby user_age.rb

# input variables
name = ""
years = 0
MONTHS_PER_YEAR = 12 # a constant

# output variable
months = 0

# processing
print "What is your name? "
name = gets.strip

print "How many years old are you? "
years = gets.strip.to_i

puts "please enter an integer" if years == 0

months = years * MONTHS_PER_YEAR

puts "#{name}, at #{years} years old, "\
"you are #{months} months old."
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Comments

0

There are several ways of doing it. If you allow for a leading sign,

x =~ /^[+-]?\d+$/

would be a possibility.

You will also have to think whether or not you allow surrounding or embedding spaces (for instance, a space between the sign and the first digit).

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0

I assume that any string value that, when converted to a float, equals an integer is to be accepted and the integer value is to be returned. Moreover, I assume integers can be entered with the "xen" (or "xEn") notation, where x is an integer or float and n is an integer.

def integer(str)
  x = Float(str) rescue nil
  return nil if x==nil || x%1 > 0
  x.to_i
end

integer("3")       #=> 3 
integer("-3")      #=> -3 
integer("3.0")     #=> 3 
integer("3.1")     #=> nil 
integer("9lives")  #=> nil 
integer("3e2")     #=> 300 
integer("-3.1e4")  #=> -31000 
integer("-30e-1")  #=> -3 
integer("3.0e-1")  #=> nil
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0

You could use Integer() to check if a string contains an integer:

Integer('1234')
#=> 1234

Integer('foo')
#=> ArgumentError: invalid value for Integer()

This could be combined with a retry:

begin
  number = Integer(gets) rescue retry
end
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