小编vic*_*dev的帖子

我有以下格式的文字:

ERR_OUT_OF_MEM, "ERR OUT OF MEM"
ERR_SOMETHING_BAD, "ERR SOMETHING BAD"

我想用下划线替换文本中带引号的所有空格:

 ERR_OUT_OF_MEM, "ERR_OUT_OF_MEM"
 ERR_SOMETHING_BAD, "ERR_SOMETHING_BAD"

我能想到的最好的正则表达式是:

\("\w\+\)\@<= 

(那里有一个空间)

但是这只能找到每个引用字符串中的第一个空格,我需要多次重复这个空格才能获得所需的效果.

有什么办法一次性完成吗？

谢谢!

我正在学习Haskell,并试图理解类型系统.

我正在尝试编写一个函数,它返回系列'Half或Three Plus One'的长度作为输入.这是我对函数的尝试,使用递归方法(该函数仅对整数输入有效):

hotpo :: (Integral a) => a->a
hotpo n = hotpoHelper n 1

hotpoHelper:: (Integral a) => a->a->a
hotpoHelper 1 n = n
hotpoHelper num count
    | even num = hotpoHelper (truncate (num/2)) (count+1)
    | otherwise = hotpoHelper (3*num+1) (count+1)

这是我尝试在GHC 6.12.3中加载此文件时得到的错误

test.hs:8:30:
    Could not deduce (RealFrac a) from the context (Integral a)
      arising from a use of `truncate' at test.hs:8:30-45
    Possible fix:
      add (RealFrac a) to the context of
        the type signature for `hotpoHelper'
    In the …