How does forever monad work?

Question

How does forever monad work? ``` forever :: (Monad m) => m a -> m b forever a = a >> forever a ``` If I write ``` main = forever $ putStrLn "SAD, I DON'T UNDERSTAND!" ``` forever gets IO (), this isn't function, how can forever repeatedly call putStrLn?

Accepted Answer

From the definition of `forever` function, you can see that it is a standard recursive function.

```
forever :: (Monad m) => m a -> m b
forever a = a >> forever a
```

There is no magic going on there. `forever` is just a recursive function. In your particular case, this is a non terminating one. But whether it becomes a terminating or non terminating depends on how the Monad is defined for that type.

Inspect the type of `>>`, we get:

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

From that you can observe the input `m a` is just ignored. Another way to think about that is that `>>` function just performs the side effect of the first parameter passed to it. In your case the `m a` will correspond to `IO ()` since that is the type of `putStrLn`.

Since IO forms a Monad, `forever` function can also act on `IO` related functions.

How does forever monad work?

Solution

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