## Question:

Consider the following `IO` code:

``````ghci> let x = return 100 :: IO Int

ghci> :t do { a <- x; print a; return 500 }
do { a <- x; print a; return 500 } :: Num b => IO b
``````

My understanding of do notation/bind is that the following signature will be enforced by the compiler:

``````ghci> :t (>>=)
(>>=) :: Monad m => m a -> (a -> m b) -> m b
``````

In the above example, my understanding is:

• `x` has type `IO Int`
• `print a` has type `IO ()`
• `return 500` has type `Num b => IO b`

It seems to me that `IO ()` does not conform to the type, `Num b => IO b`. Additionally, `IO Int` does not conform to `Num b => IO b`, as I understand.

If this observation is valid, then why does this code compile? Must not each line, i.e. `>>=`, conform to `m b`, where `m` equals IO and `b` equals `Num b => b`?

The code you posted desugars into the following.

``````x >>= (\a ->
print a >>
return 500)
``````

Or, expanding out the definition of `(>>)`

``````x       >>= (\a ->
print a >>= (\_ ->
return 500))
``````

Then, you can see that in the different calls to `(>>=)`, the types `a` and `b` are not necessarily the same. Say `(>>=) :: Monad m => m a -> (a -> m b) -> m b`.

• in the first call: `x` has type `IO Int`, `\a -> print a >>= (\_ -> return 500)` has type `Num c => Int -> IO c`, so the `a` in our type signature for `(>>=)` is `Int`, and the `b` is `c` (with the `Num` restriction).

• in the second call: `print a` has type `IO ()`, and `\_ -> return 500` has type `Num c => () -> IO c` (the `()` part is inferred from trying to match the signature of `(>>=)`) so the `a` in our type signature for `(>>=)` is `()`, and the `b` is `c` (still with the `Num` restriction).

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