## Result of expressions in Haskell with monads

I'm currently preparing for my final exam regarding Haskell, and I am going over the Monads and we were giving an example such as:

Given the following definition for the List Monad:

```
instance Monad [] where
m >>= f = concatMap f m
return x = [x]
```

where the types of `(>>=)`

and `concatMap`

are

```
(>>=) :: [a] -> (a -> [b]) -> [b]
concatMap :: (a -> [b]) -> [a] -> [b]
```

What is the result of the expression?

```
> [1,2,3] >>= \x -> [x..3] >>= \y -> return x
[1, 1, 1, 2, 2, 3] //Answer
```

Here the answer is different from what I thought it to be, now we briefly went over Monads, but from what I understand `(>>=)`

is called bind and could be read in the expression above as "applyMaybe". In this case for the first part of bind we get `[1,2,3,2,3,3]`

and we continue to the second part of the bind, but `return x`

is defined to return the list of x. Which should have been `[1,2,3,2,3,3]`

. However, I might have misunderstood the expression. Can anyone explain the wrong doing of my approach and how should I have tackled this. Thanks.

Show source

## Answers to Result of expressions in Haskell with monads ( 2 )

First, let's be clear how this expression is parsed: lambdas are

syntactic heralds, i.e. they grab as much as they can to their right, using it as the function result. So what you have there is parsed asThe inner expression is actually written more complicated than it should be.

`y`

isn't used at all, and`a >>= \_ -> p`

can just be written as`a >> p`

. There's an even better replacement though: generally, the monadic bind`a >>= \q -> return (f q)`

is equivalent to`fmap f a`

, so your expression should really be writtenor

or

At this point it should be pretty clear what the result will be, since

`>>=`

in the list monad simply maps over each element and concatenates the results.Correct.

Note that, in...

...

`x`

is bound by (the lambda of) the first`(>>=)`

, and not by the second one. Some extra parentheses might make that clearer:In

`\y -> return x`

, the values bound to`y`

(that is, the elements of`[1,2,3,2,3,3]`

) are ignored, and replaced by the corresponding values bound to`x`

(that is, the elements of the original list from which each`y`

was generated). Schematically, we have: