CODEWITHSUNDEEP
lua
user4127117
user4127117

Reputation:

lua function argument treated as string?

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!

Upvotes: 1

Views: 201

Answers (1)

Christian L.W.
Christian L.W.

Reputation: 32

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'

Upvotes: 1

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