Lua problems with variables

Question

I wrote a Lua code to arrange a list but when I enter any number with 2 decimals as 10, 20, etc. in the list, the variable 'ordenado' always takes the value 1 although whether it fulfills the conditions or not.

valor = {}
ordenado = 0

function inicializar ()
  for i = 1,10 do
    print ("Introduzca el valor "..i..":")
    valor[i] = io.read()  
  end 
end

function verificar ()
  for i = 2, #valor do
    if valor[i]>valor[i-1] then
      ordenado = ordenado + 0
    else
      ordenado = ordenado + 1
    end
    print ("actual: "..valor[i].." \nanterior: "..valor[i-1].."\nordenado:"..ordenado.."\n")
  end
end

function imprimir()
  if ordenado == 0 then
    print "La lista esta ordenada"
  else 
    print "La lista no esta ordenada"
  end
end

a = inicializar()
a = verificar()
a = imprimir()

The Lua version is 5.2.

What do you intend with a statement like ordenado = ordenado + 0? — user2033018
– user2033018, Commented Mar 12, 2014 at 15:36
Use valor[i] = tonumber(io.read()) to get numbers, not strings — Egor Skriptunoff
– Egor Skriptunoff, Commented Mar 12, 2014 at 15:38
faranwath: It is because the next function if the list is arranged then ordenado only take the value 0 and the program print that the list is arrenged. Sorry to write the code in Spanish — Dominuskernel
– Dominuskernel, Commented Mar 12, 2014 at 19:19
@EgorSkriptunoff Skriptunoff: you right, with this valor[i] = tonumber(io.read()) it work now — Dominuskernel
– Dominuskernel, Commented Mar 12, 2014 at 22:36

lhf · Accepted Answer · 2014-03-12 16:13:39Z

1

The line valor[i] = io.read() stores a string in valor[i]. As strings, "2" > "10".

answered Mar 12, 2014 at 16:13

lhf

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2 Comments

Dominuskernel Over a year ago

Won't I have this problem if I write valor[i] = tonumber(io.read())?

lhf Over a year ago

@Dominuskernel, probably not. Try it.

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