【发布时间】:2014-11-16 21:15:46
【问题描述】:
我有以下代码:
foo :: Int -> IO ()
foo n = do
x <- bar 6
print "foo running..."
print x
bar :: Int -> IO Int
bar n = do
print "bar running..."
return (n*2)
现在我想将 "x
foo :: Int -> IO ()
foo n = do
print "foo running..."
print x
where
x <- bar 6
bar :: Int -> IO Int
bar n = do
print "bar running..."
return (n*2)
我该怎么做?
【问题讨论】:
标签: haskell