lua function argument treated as string?

Question

I am trying to make this program to factor a number work in lua, and everything works except for this one line of code in it. Here's the code:

function factor(a)
 print("factoring: " .. a)
 print()
 totali = 0
 totaldiv = 0
 for i = 1, a do
  if (a%i == 0) then
   if (i<a) then
    totaldiv = totaldiv + 1
   end
   print(i)
   i = i + 1
   totali = totali + 1
  else
   i = i + 1
  end
 end
 if totali == 2 then
  print("That is a prime number!")
 elseif totaldiv == a then
  print("That is a perfect number!")
 end
end
io.write("Enter a number to factor: ")
some = io.read()
factor(some)
io.read()

The offensive line is if (i<a) then from what I've seen. What am I doing wrong? Thanks!

The value you get back from io.read will be a string, yes. You need to convert it to a number if you want that (lua will do that implicitly in some cases too). — Etan Reisner
– Etan Reisner, Commented Jul 8, 2015 at 18:22
@Etan Reisner, How do I make the number the variable a? If it was, wouldn't it output 6 as a perfect number, or am I wrong in another part of it? It seems to me it's not making a the actual function argument. I did "i<tonumber(a)" EDIT: I changed it to totaldiv = totaldiv + i, but still no luck. — user4127117
– user4127117, Commented Jul 8, 2015 at 18:27
I have no idea what problem you are actually having. Your post was cut off and you didn't say what the problem was. I just gave a generic answer about io.read and string values. Fix your post, clearly explain the exact problem and I might be able to help more. — Etan Reisner
– Etan Reisner, Commented Jul 8, 2015 at 18:31
Ok... It cut off my post for some reason. I'm trying to get my program to output "That is a perfect number!" If the sum of its proper divisors is equal to the number itself. From what I'm seeing, it's not even doing anything with that part. I tried printing totaldiv after the for loop right before the second to last end, and it doesn't even print anything... I suck at this. OK I converted the last a to a number and it worked! Thanks! — user4127117
– user4127117, Commented Jul 8, 2015 at 18:34
Replace factor(some) with factor(some + 0) or factor(tonumber(some)) — Egor Skriptunoff
– Egor Skriptunoff, Commented Jul 9, 2015 at 10:18

Christian L.W. · Accepted Answer · 2015-11-16 09:22:48Z

1

if (i<tonumber(a)) then should work.
You requested an input, which will be returned as a string.
Therefore, you can't do if (i<a) then, because you're comparing number and string via <.
You simply can't say, that 2 is less than '4'

edited Nov 16, 2015 at 9:22

answered Nov 9, 2015 at 10:36

Christian L.W.

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1 Comment

Christian L.W. Over a year ago

Yeah, sorry... I was probably tired, when writing this. It's fixed now!

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