Question

Network.accept返回一个关于接受连接的句柄，我将其作为参数传递给respondHTTP（来自Network.HTTP.Base）他们的类型是，

accept :: Socket -> IO (Handle, HostName, PortNumber)
respondHTTP :: HStream ty => HandleStream ty -> Response ty -> IO ()

这一行 - git-code

handleResponse = Kleisli (print ||| respondHTTP c)

，抛出错误，

simpleserver.hs:43:53:
    Couldn't match expected type ‘Network.TCP.HandleStream ty’
                with actual type ‘Handle’
    Relevant bindings include
      handleResponse :: Kleisli IO (Either b (Response ty)) ()
        (bound at simpleserver.hs:43:5)
    In the first argument of ‘respondHTTP’, namely ‘c’
    In the second argument of ‘(|||)’, namely ‘respondHTTP c’

在将Handle传递给Network.accept)之前，如何将Stream（从respondHTTP返回到<?php $imgfile = fopen(image/location.jpg); $loadimg = imagecreatefromjpeg($imgfile); imagejpeg($loadimg); ?>？

Answer 1

来自||的{{1}}和来自HandleStream的{{1}}是两件不同的事情;没有超类 - 子类关系。它们不可互换。

创建Network.TCP，然后将其传递给socketConnection，而不是Handle。这将返回您需要的System.IO。

Stream Vs Handle，respondHTTP抛出错误

1 个答案: