Handling of Haskell IO Actions

Asked 3 months ago, Updated 3 months ago, 12 views

I'm a beginner at Haskell.
I have a question.

parseDnsMessage::BG.BitGet DnsMessage


resolveName qname name bstr=do
  let newbstr = BSL.toStrict$replace(BS.pack qname) (BS.pack name) bstr
  retbstr<-recQuery newbstr (head rootServers 4)
  let msg = BG.runBitGet retbstr parseDnsMessage
  case msg of
    Right m->(intercalate .. $$map show(rdata$head$answer$m))

--- The following error message ---

Couldn't match expected type '[BSI.ByteString]'
            with actual type 'IO BSI.ByteString'
Inastmt of a'do' block:
  retbstr<-recQuery newbstr (head rootServers 4)
In the expression:
  do {let newbstr
             = BSL.toStrict$replace(BS.pack qname)(BS.pack name)bstr;
       retbstr<-recQuery newbstr(head rootServers4);
       let msg = BG.runBitGet retbstr parseDnsMessage;
       case msg of {
         Right m
           ->(intercalate"."$map show(rdata$head$answer$m))}

How is it appropriate to write in such a case?
Thank you for your cooperation.


2022-09-30 12:00

1 Answers

Although the do notation appears to be processed by <- while extracting the contents, there are restrictions, and the value represented by the entire do notation must be the type of monad from which the contents are taken out.

In other words, if you want to use <- to describe something, the result must also be IO.

So at least resolveName is

resolveName::[Word8]->[Word8]->BS.ByteString->IO String

It should be.In order to achieve this, I think the functions required will be those that achieve String->IO String.You can use return for this.

If you feed the return value of the last intercalate and transform it into a correct shape, you will get what you want.

2022-09-30 12:00

If you have any answers or tips

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