Implementing = in newtype

Question

Let me start from the task I want solve, probably I'm going wrong way. I use Snap framework for toy project, and the main is that it's functions under `Snap` monad. I need to add my state above it. I use monad transformer: ``` type SnapApp a = StateT AppState Snap a ``` This defined in the module, say, `Base`. Since I need it in other modules, I have to export it: ``` module Base ( .. , SnapApp ) where ``` This is good, but I want that module doesn't exported that `SnapApp` is state monad, because I have some complicated processing for setting some attributes for the state. For example, session. I have to write file when it is changed, so it wrong to just `get` and than `put` modified session, special function should be called. So, I hide that using `newtype` and not data exporting construstor: ``` newtype SnapApp a = SnapApp (StateT AppState Snap a) ``` I made it instance of my class with functions for modifying session and etc. But problem arises: I lost instances of `Monad` class a…

Accepted Answer

Let the types guide you. You need

```
(>>=) :: SnapApp a -> (a -> SnapApp b) -> SnapApp b
```

You have

```
(>>=) :: StateT AppState Snap a -> (a -> StateT AppState Snap b) -> StateT AppState Snap b
```

You need to convert:

- argument `SnapApp a` to `StateT AppState Snap a`

- argument `a -> SnapApp b` to `a -> StateT AppState Snap b`

- result `StateT AppState Snap b` to `SnapApp b`

1) use pattern matching; define:

```
fromSnapApp (SnapApp x) = x
```

2) Compose the functions `a -> SnapApp b` and `SnapApp b -> StateT AppState b`

3) Use `SnapApp`

Final result:

```
x >>= f = SnapApp (fromSnapApp x >>= (fromSnapApp . f)) 
```

or:

```
SnapApp x >>= f = SnapApp (x >>= (fromSnapApp . f)) 
```

You don't have to write this; GHC can derive the instance if you enable the `GeneralizedNewtypeDeriving` extension:

```
newtype SnapApp a = SnapApp (StateT AppState Snap a) deriving (Monad)
```

Implementing >>= in newtype

Solution

Problem