[Haskell] IORef sharing

[Jonathan Cast]

On Mon, 2008-10-27 at 17:02 -0600, Rodney D Price wrote:
> My old, deeply flawed mental picture had "iio" taking
> the role of a pointer to a value.

Not so much flawed: you just need to realize that Haskell considers the
sub-program `create a new IORef with contents 0 and return it' to be a
perfectly good value, and when you say

  iio = newIORef 0

Haskell is perfectly happy to point iio at that sub-program.

Your problem is distinguishing `program' from `value' and thinking that
`value' means `result of program'.  Haskell knows no such distinction.

> My bright, shiny
> new mental picture has "iio" acting just like a C
> #define macro:

But this is a good intuition, too.  Except without the weird syntax
bugs.  Also statically typed.  And you can use recursion, etc.

Other than that, `Haskell function = macro' isn't a bad (component of a)
mental picture.

> every time I call "iio", I'm really
> just writing "newIORef 0".

Write.  So you say

  name = expression

in Haskell when `name' is clearer in the contexts where it's used than
`expression' is (or when expression needs to be recursive).

jcc