Keep the decimal numbers?

I am making a POS system(a cash register), and for the total, I want to keep the format like 5.00, instead of 5 if the total is $5. Basically, I need help keeping a number a "Float" or "Double" type instead of having lua convert it automatically to an integer if there are no "necessary" decimals.

Look into string.format. There are some parts of it that don't work well, but I think that the decimals formatting part does work.

Lyqyd, on 07 December 2013 - 06:48 PM said:

but I think that the decimals formatting part does work.

Unfortunately I've never been able to get it to work, originally I thought it may have been 'cause of the difference between Lua and Java's string.format, but upon further investigating they're identical in how they function, but its broken, this string.format("%.2f", math.pi) outputs 3.141592653589793 which is clearly wrong.

As such you'd need to do something like this;
string.format("$%d.%.2d", dollars, cents) which when given input string.format("$%d.%.2d", 2, 5) will output $2.05.

The difference between these:

• adding the .2 in the standard floating point value format %f specifies that there needs to be two decimal places in a floating point value adding any trailing zeros where appropriate (unfortunately broken);
• adding the .2 in the standard double value format %d specifies that there needs to be a minimum of two digits in the double adding any leading zeros where appropriate. With the string.format a double does not preserve any decimal values and is cast to an integer. this can be seen by testing the following code string.format("%d", math.pi) and you'll notice that 3 is the result;

Sorry for bump.
There's a workaround to this problem

local function round(num, dec)
  local shift = 10^(dec or 2)
  return math.floor(num * shift + 0.5) / shift
end
round(math.pi) -> 3.14
round(math.pi, 4) -> 3.1416

Anavrins, on 02 December 2014 - 11:19 AM said:

Sorry for bump.
There's a workaround to this problem

local function round(num, dec)
  local shift = 10^(dec or 2)
  return math.floor(num * shift + 0.5) / shift
end
round(math.pi) -> 3.14
round(math.pi, 4) -> 3.1416

That's not the problem. He wants 4.00 to stay 4.00 and not automatically turn into 4.

That was about the problem theoriginalbit stated.
But I guess we can do something like this instead.

local function round(num, dec)
  local shift = 10^(dec or 2)
  num = math.floor(num * shift + 0.5) / shift
  if num == math.floor(num) then num = tostring(num.."."..("0"):rep(dec)) end
  return num
end

except then it returns as a string, instead of a number meaning it would no longer be an adequate round function

Normal behavior you would expect with string.format using %.2f would also return a string, my function was pretty much a work around for that string.format bug.
You would use the result of that function mainly for display purpose, not actual calculations.