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Converting numbers

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Revision as of 19:37, 20 June 2007

Converting between numerical types in Haskell must be done explicitly. This is unlike languages (such as C or Java) which automatically cast between numerical types in certain situations.

Contents

1 Converting from Integral

Integral types are ones which may only contain whole numbers and not fractions.
Int
(fixed-size machine integers) and
Integer
(arbitrary precision integers) are the two Integral types in the standard Haskell libraries. The workhorse for converting Integral types is
fromIntegral
, which will convert any integral type into any numeric type (e.g.
Rational
,
Double
...):
fromIntegral :: (Num b, Integral a) => a -> b
For example, if you have an
Int
value
n
, you cannot take its square root by typing
sqrt n
, since
sqrt
may only be applied to Floating values. Instead, you must write
sqrt (fromIntegral n)
to explicitly convert
n
to a non-integral type.

2 Converting to Rational

To convert something to a
Rational
type, you can use the function
toRational
:
toRational :: (Real a) => a -> Rational
Values of type
Rational
represent rational numbers exactly as the ratio of two
Integer
s. Applying
toRational
to an
Integral
value
n
will produce the rational number
n % 1
; applying
toRational
to a decimal (i.e.
Fractional
or
Floating
) value will produce a rational approximation. You can also construct
Rational
values explicitly using the
%
operator.

3 Converting to Integral

This is an inherently lossy transformation since integral types cannot express non-whole numbers. Depending on how you wish to convert, you might choose one of several methods.

  • ceiling
  • floor
  • truncate
  • round

4 Automatic conversion

Repeatedly people ask for automatic conversion between numbers. This is usually not a good idea; for more information, refer to the thoughts about a Generic number type.

5 original

Hi, I am trying to write some funs that convert between two coordinate systems. The first coordinate system, which ill call coord1, starts in the upper left at (0, 0) and ends in the lower right at (500, 500). Coordinates in coord1 have type (Int, Int). The second coord system, which ill call coord2, starts in the lower left at (0.0, 0.0) and ends in the upper right at (1.0, 1.0). Coords in coord2 have type (Float, Float). I was hoping someone could help me figure out how I can rewrite the two functions below so that the type checker will accept them.

 coord1ToCoord2 :: (Int, Int) -> (Float, Float)
 coord1ToCoord2 (x, y) = (x/500, (500-y)/500)
 
 coord2ToCoord1 :: (Float, Float) -> (Int, Int)
 coord2ToCoord1 (x, y) = (500/(1/x), 500 - 500/(1/y))

examples of what i want. i think i have the logic right :)

 coord1ToCoord2 (0, 0) -> (0.0, 1.0)
 coord1ToCoord2 (250, 250) -> (0.5, 0.5)
 coord1ToCoord2 (350, 350) -> (0.7, 0.3)
 coord1ToCoord2 (500, 500) -> (1.0, 0.0)
 
 coord2ToCoord1 (0.0, 0.0) -> (0, 500)
 coord2ToCoord1 (0.5, 0.5) -> (250, 250)
 coord2ToCoord1 (0.7, 0.7) -> (350, 150)
 coord2ToCoord1 (1.0, 1.0) -> (500, 0)

Ah. I realize what is messing me up.

When i saw an expression like

 500 * 0.2

i had assumed that 500 :: Integer because it didnt end in a .0. but it actually has type Double. so my problem was i would do something like this

 (toInteger (500 :: Int)) * 0.2

which of course the typechecker wouldnt accept. now that i have rid myself of my incorrect assumptions i see that i should be writing

 (fromRational (toRational (500 :: Int)) * 0.2) :: Float

now that i have a better understanding i am able to write my funs. thank you for your help :)