# Printing the bits of an Int|Double Classic List Threaded 4 messages Open this post in threaded view
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## Printing the bits of an Int|Double

 I've been trying to play around with binary data, but I haven't made much progress trying to print the bits of a Double. With help from #haskell I've made it this far: -- Printing the bits of an Int main = do putStrLn \$ showIntAtBase 2 (chr . (48+)) z "" -- 103 = 1100111, after bit shifting 11001 z = shiftR (103 :: Int64) 2 This is as far as I got with Doubles: import Data.Binary.IEEE754 import qualified Data.ByteString.Lazy as BS main = do BS.putStrLn \$ runPut \$ putFloat64be 4.123 Instead of playing with ByteStrings, is there just a way to fill an Int64 with the bits of a Double (Similar to Java's long = Double.doubleToLongBits(double))?       -------------- next part -------------- An HTML attachment was scrubbed... URL: http://www.haskell.org/pipermail/beginners/attachments/20091211/4f107c25/attachment.html
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## Printing the bits of an Int|Double

 Hello M Xyz Here's a module that seems to work for single precision floats - you might want to do some checking of random numbers with toAndFro (there's plenty of room for errors). The print functions is printBin uses showIntAtBase which doesn't pad with leading zeros, you might want to add them. Best wishes Stephen  I wrote it a while ago but checking this morning it had some nefarious bugs in. module PrintIeeeFloat where import Data.Bits import Data.Char import Data.Word import Numeric const_B :: Int const_B = 127 printBin :: (Fractional a,Ord a) => a -> ShowS printBin a =     f s . showChar ' ' . f t . showChar ' ' . f u . showChar ' ' . f v   where     f = showIntAtBase 2 (chr . (48+))     (s,t,u,v) = packIEEESingle a toAndFro :: (Fractional a, Ord a) => a -> a toAndFro a = let (s,t,u,v) = packIEEESingle a in unpackIEEESingle s t u v unpackIEEESingle :: Fractional a => Word8 -> Word8 -> Word8 -> Word8 -> a unpackIEEESingle b24_31 b16_23 b8_15 b0_7 =  sign \$ fract * (2 ^^ expo)   where     sign  = if b24_31 `testBit` 7 then negate else id     expo  = exponent' b24_31 b16_23     fract = fraction b16_23 b8_15 b0_7 exponent' :: Word8  -> Word8 -> Int exponent' a b = (a' `shiftL` 1) + (b' `shiftR` 7) - 127   where     a' = fromIntegral \$ (a .&. 0x7f)     b' = fromIntegral \$ (b .&. 0x80) fraction :: Fractional a => Word8 -> Word8 -> Word8 -> a fraction b16_24 b8_15 b0_7 = 1.0 + ((fromIntegral frac) / (2 ^^ 23))   where    frac :: Int    frac = (shiftL16 (b16_24 .&. 0x7f)) + (shiftL8 b8_15) + fromIntegral b0_7 packIEEESingle :: (Fractional a,Ord a) => a -> (Word8,Word8,Word8,Word8) packIEEESingle a = (flipSign b24_31, exp_part+mant_part, b8_15, b0_7)   where     k     = findPosExpo \$ abs a     e     = k + const_B     halfa = (abs a) / (2 ^^ fromIntegral k)     f     = expand \$ halfa - 1     (b24_31, exp_part)          = expoWords e     (mant_part,b8_15, b0_7)     = mantWords f     flipSign = if a > 0 then id else (`setBit` 7) findPosExpo :: (Fractional a, Ord a) => a -> Int findPosExpo r | r <= 0    = 0               | otherwise = step r 1   where     step r' k | r <= fromIntegral 2 ^^ k = k-1               | otherwise                = step r' (k+1) expand :: (Fractional a, Ord a) => a -> Word32 expand n = (`shiftR` 9) \$ step n 0 id   where     step x ix f | x <= 0    = f (0::Word32)                 | otherwise = let y = 1 / (2 ^^ (ix+1))                               in if x >= y                                  then step (x-y) (ix+1) (f . (`setBit` (31-ix)))                                  else step x (ix+1) f -- 7 bits left, 1 bit right expoWords :: Int -> (Word8,Word8) expoWords n = (left, right)   where     right = if n `testBit` 0 then 128 else 0     left  = fromIntegral \$ n `shiftR` 1 mantWords :: Word32 -> (Word8,Word8,Word8) mantWords x = (a,b,c)   where     c = fromIntegral \$ x .&. 0xff     b = fromIntegral \$ (`shiftR` 8)  \$ x .&. 0xff00     a = fromIntegral \$ (`shiftR` 16) \$ x .&. 0xff0000 shiftL8 :: (Bits b, Integral b) => Word8 -> b shiftL8 = (`shiftL` 8) . fromIntegral shiftL16 :: (Bits b, Integral b) => Word8 -> b shiftL16 = (`shiftL` 16) . fromIntegral shiftL24 :: (Bits b, Integral b) => Word8 -> b shiftL24 = (`shiftL` 24) . fromIntegral w32be :: Word8 -> Word8 -> Word8 -> Word8 -> Word32 w32be a b c d = (shiftL24 a) + (shiftL16 b) + (shiftL8 c) + fromIntegral d 2009/12/11 M Xyz <[hidden email]> > > I've been trying to play around with binary data, but I haven't made much progress > trying to print the bits of a Double. With help from #haskell I've made it this far: > > -- Printing the bits of an Int > main = do putStrLn \$ showIntAtBase 2 (chr . (48+)) z "" > > -- 103 = 1100111, after bit shifting 11001 > z = shiftR (103 :: Int64) 2 > > This is as far as I got with Doubles: > > import Data.Binary.IEEE754 > import qualified Data.ByteString.Lazy as BS > main = do BS.putStrLn \$ runPut \$ putFloat64be 4.123 > > Instead of playing with ByteStrings, is there just a way to fill an Int64 with the bits of a Double (Similar to Java's long = Double.doubleToLongBits(double))? > > > _______________________________________________ > Beginners mailing list > [hidden email] > http://www.haskell.org/mailman/listinfo/beginners>