On Fri, Jun 22, 2012 at 11:18 AM, Captain Freako wrote:
> Okay, I get it now. Thanks to all of you for your quick replies!
>
> So, here's what I need to do:
>
> My Haskell code gets called by a higher level C function and asked to
> initialize itself.
> As part of that initialization, I'm expected
Okay, I get it now. Thanks to all of you for your quick replies!
So, here's what I need to do:
1. My Haskell code gets called by a higher level C function and asked to
initialize itself.
2. As part of that initialization, I'm expected to return a pointer to
an instance of a particular
On Fri, Jun 22, 2012 at 5:30 PM, Captain Freako wrote:
> 12 main = do
> 13 let theValue = 1
> 14 print theValue
> 15 theValueRef <- newIORef theValue
> 16 bump theValueRef
> 17 return theValue
theValue is a plain old immutable Haskell variable. "newIORef" creates
an IORe
theValueRef isn't a pointer to theValue that you can use to somehow change
theValue (which is immutable).
theValueRef is a reference to a "box" that contains a totally separate,
mutable value.
When you use newIORef to create theValueRef, it's *copying* theValue into
the box. When you mutate theVal
Hi experts,
I fear I don't understand how to properly use *Data.IORef*.
I wrote the following code:
1 -- Testing Data.IORef
2 module Main where
3
4 import Data.IORef
5
6 bump :: IORef Int -> IO()
7 bump theRef = do
8 tmp <- readIORef theRef
9 let tmp2 = tmp + 1
10
