So how do I force IO actions whose results are discarded (including IO ()) to
be strict?
main = do
s<-newIORef (1::Int)
let
f :: Int -> Int -> IO Int
f 0 !acc = return acc -- note strict accumulator
f n !acc = do
v <- modifyIORef s (+2) >>readIORef s -- reading
immediately after writing
f (n-1) (v+acc)
f 1000000 100 >>= print
readIORef s>>=print
runs OK, while
main = do
s<-newIORef (1::Int)
let
f :: Int -> Int -> IO Int
f 0 !acc = return acc -- note strict accumulator
f n !acc = do
v <- modifyIORef s (+2) >>return 1
f (n-1) (v+acc)
f 1000000 100 >>= print
readIORef s>>=print
,
main = do
s<-newIORef (1::Int)
let
f :: Int -> Int -> IO Int
f 0 !acc = return acc -- note strict accumulator
f n !acc = do
v <- modifyIORef s (+2) >>readIORef s>>return 1
f (n-1) (v+acc)
f 1000000 100 >>= print
readIORef s>>=print
and
main = do
s<-newIORef (1::Int)
let
f :: Int -> Int -> IO Int
f 0 !acc = return acc -- note strict accumulator
f n !acc = do
v <- (>>return 1) $! (modifyIORef s (+2) >>readIORef s)
f (n-1) (v+acc)
f 1000000 100 >>= print
readIORef s>>=print
all overflows after correctly printing the first number
----- 原始邮件 -----
发件人: "Johan Tibell" <[email protected]>
收件人: [email protected]
抄送: [email protected]
发送时间: 星期四, 2012年 9 月 20日 上午 1:28:47
主题: Re: [Haskell-cafe] How to implement nested loops with tail recursion?
On Wed, Sep 19, 2012 at 7:24 PM, <[email protected]> wrote:
> main = do
> let
> f 0 acc = return acc
> f n acc = do
> v <- return 1
> f (n-1) (v+acc)
> f 1000000 100 >>= print
Try this
main = do
let
f :: Int -> Int -> IO Int
f 0 !acc = return acc -- note strict accumulator
f n acc = do
v <- return 1
f (n-1) (v+acc)
f 1000000 100 >>= print
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