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99 lines
2.7 KiB
Haskell
99 lines
2.7 KiB
Haskell
{-|
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a simple amount is a currency, quantity pair:
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$1
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£-50
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EUR 3.44
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GOOG 500
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1.5h
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90m
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0
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a mixed amount is one or more simple amounts:
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$50, EUR 3, AAPL 500
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16h, $13.55, oranges 6
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currencies may be convertible or not (eg, currencies representing
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non-money commodities). A mixed amount containing only convertible
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currencies can be converted to a simple amount.
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arithmetic:
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$1 - $5 = $-4
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$1 + EUR 0.76 = $2
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EUR0.76 + $1 = EUR 1.52
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EUR0.76 - $1 = 0
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($5, 2h) + $1 = ($6, 2h)
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($50, EUR 3, AAPL 500) + ($13.55, oranges 6) = $67.51, AAPL 500, oranges 6
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($50, EUR 3) * $-1 = $-53.96
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($50, AAPL 500) * $-1 = error
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-}
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module Amount
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where
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import Utils
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import Types
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import Currency
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tests = runTestTT $ test [
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show (dollars 1) ~?= "$1.00"
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,show (hours 1) ~?= "1h" -- currently h1.00
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,parseAmount "$1" ~?= dollars 1 -- currently 0
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]
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nullamt = dollars 0
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parseAmount :: String -> Amount
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parseAmount s = nullamt
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instance Show Amount where show = showAmountRounded
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showAmountRounded :: Amount -> String
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showAmountRounded (Amount c q p) =
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(symbol c) ++ ({-punctuatethousands $ -}printf ("%."++show p++"f") q)
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showAmountRoundedOrZero :: Amount -> String
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showAmountRoundedOrZero a@(Amount c _ _) =
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let s = showAmountRounded a
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noncurrency = drop (length $ symbol c)
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nonnulls = filter (flip notElem "-+,.0")
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iszero = (nonnulls $ noncurrency s) == ""
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in if iszero then "0" else s
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punctuatethousands :: String -> String
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punctuatethousands s =
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sign ++ (punctuate int) ++ frac
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where
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(sign,num) = break isDigit s
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(int,frac) = break (=='.') num
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punctuate = reverse . concat . intersperse "," . triples . reverse
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triples "" = []
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triples s = [take 3 s] ++ (triples $ drop 3 s)
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instance Num Amount where
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abs (Amount c q p) = Amount c (abs q) p
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signum (Amount c q p) = Amount c (signum q) p
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fromInteger i = Amount (getcurrency "$") (fromInteger i) amtintprecision
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(+) = amountop (+)
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(-) = amountop (-)
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(*) = amountop (*)
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-- | problem: when an integer is converted to an amount it must pick a
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-- precision, which we specify here (should be infinite ?). This can
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-- affect amount arithmetic, in particular the sum of a list of amounts.
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-- So, we may need to adjust the precision after summing amounts.
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amtintprecision = 2
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-- | apply op to two amounts, adopting a's currency and lowest precision
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amountop :: (Double -> Double -> Double) -> Amount -> Amount -> Amount
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amountop op (Amount ac aq ap) b@(Amount _ _ bp) =
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Amount ac (aq `op` (quantity $ toCurrency ac b)) (min ap bp)
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toCurrency :: Currency -> Amount -> Amount
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toCurrency newc (Amount oldc q p) =
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Amount newc (q * (conversionRate oldc newc)) p
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