问题
I made really time consuming algorithm which produces a short string as the result. When I try to print it (via putStrLn) it appears on the screen character by character. I did understand why that happened, and I tried to force evaluation of the string before actual printing.
myPrint !str = putStrLn str
But this help very little. When I ran the program in debug I noticed that the !str forced evaluation only for the first character.
Does anyone know why is that, and how to deal with this?
回答1:
(!) translates into seq, which evaluates strictly to Weak Head Normal Form -- that is, it only evaluates to the outermost constructor. To evaluate more deeply, you need a "deep" form of seq.
This is known as deepseq.
It is in the deepseq package.
回答2:
seqList :: [a] -> ()
seqList [] = ()
seqList (x:xs) = strictList xs
来源：https://stackoverflow.com/questions/5558043/forced-strictness-for-lists-in-haskell