了解`评估'功能
Haskell文档解释了evaluate函数：
当生成的IO操作执行时，强制其参数被评估为弱头标准形式。
Prelude Control.Exception> let xs = [1..100] :: [Int] Prelude Control.Exception> :sprint xs
xs = _
Prelude Control.Exception> let ys = evaluate xs
Prelude Control.Exception> :t ys
ys :: IO [Int]
Prelude Control.Exception> ys
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,6Prelu2,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100]
Prelude Control.Exception> :sprint xs
xs = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,
24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,
46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,
68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,
90,91,92,93,94,95,96,97,98,99,100]
Prelude Control.Exception> :sprint ys
ys = _
为什么ys不是弱头正常形式，即:sprint ys不等于_ : _ ？
你的ys值有一个IO [Int]类型。 现在，IO是一种抽象类型，在你的情况下可以被认为是RealWorld -> ([Int], RealWorld) 。 现在这个IO值已经处于弱首标范围。 这就是为什么当你做sprint时你会将它看作是_的原因。
为什么ys不是弱头正常形式，即:sprint ys不等于_ : _ ？
ys外部术语不能是_ : _因为它不是一个列表，但它是IO [Int]类型的值。
除了Sibi所说的之外，这里还有一种方法可以看到evaluate实际上完成了文档所说的内容：
GHCi> let xs = [1..100] :: [Int]
GHCi> :sprint xs
xs = _
GHCi> let a = evaluate xs >> return ()
GHCi> a
GHCi> :sprint xs
xs = 1 : _