Question

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

使用严格字段（!Int），我认为会立即评估undefined，这会导致异常，但实际上，在打印之前它仍然没有评估。那是为什么？

> 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 - let从不自行评估任何内容（除非使用BangPatterns）。

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

你可以区分严格和懒惰的构造函数，如：

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

Haskell严格的字段

1 个答案: