Haskell 严格字段

Question

定义惰性字段时，没有异常(exception)，直到您打印它。

> data T = T Int deriving (Show)
> let t = T undefined
> t
T *** Exception: Prelude.undefined
CallStack (from HasCallStack):
  error, called at libraries/base/GHC/Err.hs:79:14 in base:GHC.Err
  undefined, called at <interactive>:3:7 in interactive:Ghci3

对于严格的领域( !Int )，我认为 undefined将立即评估，这将导致异常，但实际上，在您打印它之前它仍然没有评估。这是为什么？

> data X = X !Int deriving (Show)
> let x = X undefined
> x
*** Exception: Prelude.undefined
CallStack (from HasCallStack):
  error, called at libraries/base/GHC/Err.hs:79:14 in base:GHC.Err
  undefined, called at <interactive>:6:11 in interactive:Ghci5

Answer 1

因为let本身定义了一个惰性绑定(bind)——let从不单独评估任何东西(除非使用 BangPatterns)。

ghci> let x = undefined
ghci> x
*** Exception: Prelude.undefined

您可以像这样区分严格构造函数和惰性构造函数:

ghci> T undefined `seq` ()
()
ghci> X undefined `seq` ()
*** Exception: Prelude.undefined