Unwrap value from IO operation at a later time

Hello i was wondering how can you unwrap a value at a later time in the IO monad? If a<-expression binds the result to a then can't i use (<-expression) as a parameter for a given method eg: method (<-expression) where method method accepts the result of the evaluation?

Code

let inh=openFile "myfile" WriteMode
let outh=openFile "out.txt" WriteMode 
hPutStrLn (<-outh) ((<-inh)>>=getLine)

I have not entered the Monad chapter just basic <- and do blocks but i suppose it has to do with monads. Then if i want to pass the result if the evaluation to hGetLine can't i use something like:

(<-expression)=>>hGetLine

Solution

You already understand that <- operator kind of unwraps IO value, but it's actually the syntax of do notation and can be expressed like this (actually I'm not sure, which results you're trying to achieve, but the following example just reads the content from one file and puts the content to another file):

import System.IO

main = do
  inh <- openFile "myfile" ReadMode
  outh <- openFile "out.txt" WriteMode
  inContent <- hGetLine inh
  hPutStrLn outh inContent
  hClose outh

According to documentation hGetLine, hPutStrlLn and hClose accept values of Handle type as an argument, but openFile returns IO Handle, so we need to unwrap it using <- operator

But if you want to use >>= function instead, then this is one of the options of doing it:

import System.IO

writeContentOfMyFile :: Handle -> IO ()
writeContentOfMyFile handler =
  openFile "myfile" ReadMode >>= hGetLine >>= hPutStrLn handler

main =
  withFile "out.txt" WriteMode writeContentOfMyFile