App Store上的应用程序Snapchat是一款应用程序,可让您在其上分享带有自毁功能的图片.您只能查看X秒的照片.如果您在使用家用电源键组合显示图片时尝试截取屏幕截图,它会告诉发件人您尝试截取屏幕截图.
SDK的哪个部分可以让您检测到用户正在截取屏幕截图？我不知道这是可能的.
我正在努力GHC开发Ubuntu.做了以下事情:
sudo apt-get install ghc
sudo apt-get install cabal-install
cabal update
cabal install hsenv
然后我尝试创建一个hsenv环境并得到以下内容:
xx@xx-VirtualBox:~/scm/t1$ hsenv
Creating Virtual Haskell directory structure
Installing GHC
Initializing GHC Package database at /home/xx/scm/t1/.hsenv/ghc_pkg_db
Copying necessary packages from original GHC package database
Failed to copy optional package ghc-binary from system's GHC:
/usr/bin/ghc-pkg process failed with status 1
Using user-wide (~/.cabal/packages) Hackage download cache directory
Installing cabal config at /home/xx/scm/t1/.hsenv/cabal/config
Installing activate script
Installing cabal wrapper using /home/xx/scm/t1/.hsenv/cabal/config at /home/xx/scm/t1/.hsenv/bin/cabal
Skipping …Control.Monad.Free 实现一个免费的monad:
data Free f a = Pure a | Free (f (Free f a))
instance Functor f => Functor (Free f) where
fmap f = go where
go (Pure a) = Pure (f a)
go (Free fa) = Free (go <$> fa)
我在理解第二go行时遇到了很多麻烦,特别是在描述自由monad是什么的情况下.有些人可以描述一下这是如何工作的以及为什么它会成为Free f a一个免费的monad？
与我得到的结果有点混淆.使用以下内容:
GHCi, version 7.4.2: http://www.haskell.org/ghc/ :? for help
Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
Prelude> :m + Language.Haskell.TH
我得到了这个成功的结果:
Prelude Language.Haskell.TH> runQ [d| data X = X |]
[DataD [] X_0 [] [NormalC X_1 []] []]
但我得到这个错误:
Prelude Language.Haskell.TH> runQ [d| data X = X deriving Show |]
<interactive>:4:30:
The exact Name `X' is not in scope
Probable cause: you used a unique name (NameU) …我在模块中有以下代码:
{-# LANGUAGE TemplateHaskell #-}
module Alpha where
import Language.Haskell.TH
import Data.List
data Alpha = Alpha { name :: String, value :: Int } deriving (Show)
findName n = find ((== n) . name)
findx obj = sequence [valD pat bod []]
where
nam = name obj
pat = varP (mkName $ "find" ++ nam)
bod = normalB [| findName nam |]
然后我在主文件中有以下内容:
{-# LANGUAGE TemplateHaskell #-}
import Alpha
one = Alpha "One" 1
two = Alpha "Two" 2
three …我正在努力让下面的代码工作.它是一个有限状态机,我传入一个函数作为下一个状态.调用该函数r并返回结果列表+下一个函数作为下一个状态.继续调用直到列表用完,并返回结果的串联.monad是一个错误monad,允许我在需要时抛出错误.
fsm f [] = return []
fsm f (r:rs) = do
(xs, f') <- f r
rest <- fsm f' rs
return $ xs ++ rest
错误是:
Occurs check: cannot construct the infinite type: t1 = t0 -> m0 ([a0], t1)
In the first argument of `fsm', namely f'
我之前看过无限类型错误,我理解它的方法是用a包装一个类型newtype.但我无法弄清楚如何完成这项工作.
有人能指出洞察力吗？
class SumRes r where
sumOf :: Integer -> r
instance SumRes Integer where
sumOf = id
instance (Integral a, SumRes r) => SumRes (a -> r) where
sumOf x = sumOf . (x +) . toInteger
对此:
class SumRes r where
sumOf :: Int -> r
instance SumRes Int where
sumOf = id
instance (SumRes r) => SumRes (Int -> r) where
sumOf x = sumOf . (x +)
我得到了Illegal instance declaration for SumRes …
我在文件中有以下内容:
import Control.Monad
ema a = scanl1 $ \m n -> (1-a)*m + a*n
macd = ema 9 . uncurry (zipWith (-)) . liftM2 (,) (ema 26) (ema 12)
在编译时,我得到以下内容:
:t macd
macd :: [Integer] -> [Integer]
然而,
:t ema 9 . uncurry (zipWith (-)) . liftM2 (,) (ema 26) (ema 12)
ema 9 . uncurry (zipWith (-)) . liftM2 (,) (ema 26) (ema 12)
:: Num a => [a] -> [a]
那么,为什么在更受限制的类型中存在差异macd呢？
在以下代码中:
a = b = c == 1
我想只匹配前两个=,而不是最后的==.
我认为该模式\<=\>将起作用,因为\<匹配开头的单词并\>匹配结束.但事实并非如此.这种模式有什么问题,正确的模式是什么？
注意:完整的源代码在这里:https://gist.github.com/anonymous/7085509
我有以下功能:
tournament n p pop = do
winner <- (\w -> min (n - 1) (floor (log w / log (1-p)))) <$> gaRandom
(flip S.index) winner <$> S.sort <$> seqChoose n pop
没有类型签名,编译器告诉我tournament签名是:
tournament
:: (Floating a, Ord a1, RealFrac a, Random a) =>
Int -> a -> S.Seq a1 -> StateT GA Data.Functor.Identity.Identity a1
哪个看起来很好.但是当我使用它时:
t2 = do
g <- newStdGen
let a = evalState (tournament 5 0.9 (S.fromList [1..10])) (GA g)
return …