Haskell严格的字段

Posted 2021-03-31

定义惰性字段时，在打印之前没有异常。

> 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

有了严格的字段（!Int），我认为undefined会被立即评估，这将导致异常，但实际上，它仍然没有评估，直到你打印它。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

答案

因为let本身定义了一个懒惰的绑定 - let从不评估任何东西（除非使用BangPatterns）。

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

您可以区分严格和惰性构造函数，如下所示：

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