Question on proper use of Data.IORef

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 writeIORef theRef tmp2 11 12 main = do 13 let theValue = 1 14 print theValue 15 theValueRef <- newIORef theValue 16 bump theValueRef 17 return theValue and got this, in ghci: *Main> :load test2.hs [1 of 1] Compiling Main ( test2.hs, interpreted ) Ok, modules loaded: Main. *Main> main 1 *1* I was expecting this: *Main> :load test2.hs [1 of 1] Compiling Main ( test2.hs, interpreted ) Ok, modules loaded: Main. *Main> main 1 *2* Can anyone help me understand what I'm doing wrong? Thanks! -db

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 theValueRef, you're mutating the value inside the
box - theValue remains unchanged.
Cheers,
Adam
On 22 June 2012 11:30, Captain Freako
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 writeIORef theRef tmp2 11 12 main = do 13 let theValue = 1 14 print theValue 15 theValueRef <- newIORef theValue 16 bump theValueRef 17 return theValue
and got this, in ghci:
*Main> :load test2.hs [1 of 1] Compiling Main ( test2.hs, interpreted ) Ok, modules loaded: Main. *Main> main 1 *1*
I was expecting this:
*Main> :load test2.hs [1 of 1] Compiling Main ( test2.hs, interpreted ) Ok, modules loaded: Main. *Main> main 1 *2*
Can anyone help me understand what I'm doing wrong?
Thanks! -db
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On Fri, Jun 22, 2012 at 5:30 PM, Captain Freako
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 IORef whose initial value is equal to the argument; it doesn't create a pointer to the value or something like that. Change "return theValue" to "readIORef theValueRef" to extract the changed value in the IORef. --Max

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 data structure, which gets created/initialized
in Haskell.
3. I create a stable pointer to a data structure instance, using *
newStablePtr*, and return that pointer to the calling C function.
4. The same C function then, at a later time, calls a different function
in my Haskell code, sending it that very same pointer, and expects that
code to modify the data structure pointed to, in place.
So, I wrote some code that looks like this:
secondFunction :: StablePtr DataStructure -> IO ()
secondFunction dsPtr = do
ds <- deRefStablePtr dsPtr
theRef <- newIORef ds
writeIORef theRef newDs
which, of course, didn't work, because I didn't understand the true nature
of IORef.
*So, my question is: How do I do this? That is, how do I modify, in place,
a data structure, which I originally created and made stable, using a
pointer to that structure, which is being passed back to me, by the higher
level C function that is calling me?*
Thanks, all!
-db
On Fri, Jun 22, 2012 at 8:43 AM, Max Rabkin
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 IORef whose initial value is equal to the argument; it doesn't create a pointer to the value or something like that. Change "return theValue" to "readIORef theValueRef" to extract the changed value in the IORef.
--Max

On Fri, Jun 22, 2012 at 11:18 AM, Captain Freako
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 to return a pointer to an instance of a particular data structure, which gets created/initialized in Haskell. I create a stable pointer to a data structure instance, using newStablePtr, and return that pointer to the calling C function. The same C function then, at a later time, calls a different function in my Haskell code, sending it that very same pointer, and expects that code to modify the data structure pointed to, in place.
So, I wrote some code that looks like this:
secondFunction :: StablePtr DataStructure -> IO () secondFunction dsPtr = do ds <- deRefStablePtr dsPtr theRef <- newIORef ds writeIORef theRef newDs
which, of course, didn't work, because I didn't understand the true nature of IORef.
So, my question is: How do I do this? That is, how do I modify, in place, a data structure, which I originally created and made stable, using a pointer to that structure, which is being passed back to me, by the higher level C function that is calling me?
You could use a StablePtr (IORef DataStructure).
Thanks, all! -db
On Fri, Jun 22, 2012 at 8:43 AM, Max Rabkin
wrote: 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 IORef whose initial value is equal to the argument; it doesn't create a pointer to the value or something like that. Change "return theValue" to "readIORef theValueRef" to extract the changed value in the IORef.
--Max
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participants (4)
-
Adam Smith
-
Antoine Latter
-
Captain Freako
-
Max Rabkin