Haskell如何確定隨機生成的數字是什麼樣的布爾值？

Question

考慮下面的兩組代碼：

random (mkStdGen 1) :: (Int, StdGen) 
-- returns (7918028818325808681,545291967 2103410263) 
random (mkStdGen 1) :: (Bool, StdGen) 
-- returns (True,80028 40692) 


random (mkStdGen 949488) :: (Int, StdGen) 
-- returns (9159618695640234475,587416689 2103410263) 
random (mkStdGen 949488) :: (Bool, StdGen) 
-- returns (False,1485632275 40692) 

爲什麼7918028818325808681轉化爲True但9159618695640234475轉化爲False？

Answer 1

Instance Bool與Instance Int共享執行，但是共享的代碼是randomR的代碼，它使用一個範圍。我們可以驗證這一點使用快速檢查：

Prelude> import Test.QuickCheck 
Prelude Test.QuickCheck> import System.Random 
Prelude Test.QuickCheck System.Random> :{ 
Prelude Test.QuickCheck System.Random| prop seed = let 
Prelude Test.QuickCheck System.Random| gen = mkStdGen seed 
Prelude Test.QuickCheck System.Random| b = fst (random gen) 
Prelude Test.QuickCheck System.Random| i = fst (randomR (0,1) gen) 
Prelude Test.QuickCheck System.Random| in if b then i == 1 else i == 0 
Prelude Test.QuickCheck System.Random| :} 
Prelude Test.QuickCheck System.Random> quickCheck prop 
+++ OK, passed 100 tests. 

您也可以看看definition of the instance Random Bool在那裏你會發現這個代碼：

instance Random Bool where 
    randomR (a,b) g = 
     case (randomIvalInteger (bool2Int a, bool2Int b) g) of 
     (x, g') -> (int2Bool x, g') 
     where 
     bool2Int :: Bool -> Integer 
     bool2Int False = 0 
     bool2Int True = 1 

    int2Bool :: Int -> Bool 
    int2Bool 0 = False 
    int2Bool _ = True 

    random g = randomR (minBound,maxBound) g 

如此重要，你在呼喚randomR (0,1)，然後映射0到False和1到True：

> random (mkStdGen 949488) :: (Bool, StdGen) 
(False,1485632275 40692) 
> randomR (0,1) (mkStdGen 949488) :: (Int, StdGen) 
(0,1485632275 40692) 
> random (mkStdGen 1) :: (Bool, StdGen) 
(True,80028 40692) 
> randomR (0,1) (mkStdGen 1) :: (Int, StdGen) 
(1,80028 40692)