Question

Give a definition of the function
fmap :: (a->b) -> IO a -> IO b

the effect of which is to transform an interaction by applying the function to its result. you should define it using the do construct.

how should I define the fmap? I have no idea about it?

could someone help me with that?

Thanks~!

Answer 1

It looks like homework or something so I will give you enough hint so that you can work rest of the details yourself.

fmap1 :: (a -> b) -> IO a -> IO b 
fmap1 f action = 

action is as IO action and f is a function from a to b and hence type a -> b.

If you are familiar with monadic bind >>= which has type (simplified for IO monad)

(>>=) :: IO a -> (a -> IO b) -> IO b

Now if you look at

action >>= f

It means perform the IO action which returns an output (say out of type a) and pass the output to f which is of type a -> IO b and hence f out is of type IO b.

If you look at the second function called return which has type (again simlified for IO monad)

return :: a -> IO a

It takes a pure value of type a and gives an IO action of type IO a.

Now lets look back to fmap.

fmap1 f action 

which performs the IO action and then runs f on the output of the action and then converts the output to another IO action of type IO b. Therefore

fmap1 f action = action >>= g 
    where
        g out = return (f out)

Now comes the syntactic sugar of do notation. Which is just to write bind >>= in another way.

In do notation you can get the output of an action by

out <- action 

So bind just reduces to

action >>= f = do 
    out <- action 
    f out 

I think now you will be able to convert the definition of fmap to do construct.