Hi

I'm stuggling with immutable.

I have a fixed size buffer which is used as a circular buffer of floats and is effectively double buffering data I wish to transfer to another thread. At an appropriate point I take the top half or bottom half of the buffer and send it to another thread.

To do this I need to copy the data to an immutable transfer buffer to use in the send. I'm sure this is simple but I can't figure it.

if I say something like:
float[] xfer = new float[512];
xfer = buffer[0 .. $/2];
tid.send(xfer);

it rightly tells me 'thread local data not allowed'. If I make it:

immutable (float)[] xfer;
xfer = buffer[0 .. $/2];
tid.send(xfer);

it tells me 'can't implicitly convert float[] to immutable (float)[]'

If I try a float by float copy into xfer it can't because I've said the buffer is immutable. In the first example I can't figure out how to convert the slice into an immutable copy which I think is what I should be doing.

Can someone point me in the right direction.

Thanks
Bob

Bob Cowdery <bob@bobcowdery.plus.com> wrote:

> if I say something like:
> float[] xfer = new float[512];
> xfer = buffer[0 .. $/2];
> tid.send(xfer);
>
> it rightly tells me 'thread local data not allowed'. If I make it:
>
> immutable (float)[] xfer;
> xfer = buffer[0 .. $/2];
> tid.send(xfer);
>
> it tells me 'can't implicitly convert float[] to immutable (float)[]'
>
> If I try a float by float copy into xfer it can't because I've said the
> buffer is immutable. In the first example I can't figure out how to
> convert the slice into an immutable copy which I think is what I should
> be doing.
>
> Can someone point me in the right direction.

http://www.digitalmars.com/d/2.0/phobos/std_exception.html#assumeUnique

float[] xfer = new float[512];
xfer = buffer[0 .. $/2];
tid.send(assumeUnique(xfer));

This should solve your problems.

-- 
Simen

 Thanks, that at least builds now. I didn't see that trick in the book
but I've found it on the on-line library ref under std.exception. I
can't say I understand why its under std.exception.

bob

On 21/09/2010 08:48, Simen kjaeraas wrote:
> Bob Cowdery <bob@bobcowdery.plus.com> wrote:
>
>> if I say something like:
>> float[] xfer = new float[512];
>> xfer = buffer[0 .. $/2];
>> tid.send(xfer);
>>
>> it rightly tells me 'thread local data not allowed'. If I make it:
>>
>> immutable (float)[] xfer;
>> xfer = buffer[0 .. $/2];
>> tid.send(xfer);
>>
>> it tells me 'can't implicitly convert float[] to immutable (float)[]'
>>
>> If I try a float by float copy into xfer it can't because I've said the buffer is immutable. In the first example I can't figure out how to convert the slice into an immutable copy which I think is what I should be doing.
>>
>> Can someone point me in the right direction.
>
> http://www.digitalmars.com/d/2.0/phobos/std_exception.html#assumeUnique
>
> float[] xfer = new float[512];
> xfer = buffer[0 .. $/2];
> tid.send(assumeUnique(xfer));
>
> This should solve your problems.
>

Bob Cowdery <bob@bobcowdery.plus.com> wrote:

>  Thanks, that at least builds now. I didn't see that trick in the book
> but I've found it on the on-line library ref under std.exception. I
> can't say I understand why its under std.exception.

I know. It used to be std.contracts, which makes a bit more sense. Then,
someone decided std.contracts should be all about exceptions, and
apparently forgot this exception.

-- 
Simen

On 09/21/2010 09:29 AM, Bob Cowdery wrote:
>   Hi
>
> I'm stuggling with immutable.
>
> I have a fixed size buffer which is used as a circular buffer of floats
> and is effectively double buffering data I wish to transfer to another
> thread. At an appropriate point I take the top half or bottom half of
> the buffer and send it to another thread.
>
> To do this I need to copy the data to an immutable transfer buffer to
> use in the send. I'm sure this is simple but I can't figure it.
>
> if I say something like:
> float[] xfer = new float[512];
> xfer = buffer[0 .. $/2];
> tid.send(xfer);
>
> it rightly tells me 'thread local data not allowed'. If I make it:
>
> immutable (float)[] xfer;
> xfer = buffer[0 .. $/2];
> tid.send(xfer);
>
> it tells me 'can't implicitly convert float[] to immutable (float)[]'
>
> If I try a float by float copy into xfer it can't because I've said the
> buffer is immutable. In the first example I can't figure out how to
> convert the slice into an immutable copy which I think is what I should
> be doing.
>
> Can someone point me in the right direction.
>
> Thanks
> Bob
>

immutable(float)[] xfer = buffer[0 .. $/2].idup;
tid.send(xfer);

Don't use assumeUnique for non-unique references.

 Ok, so assumeUnique when I read it a bit more will still share data but
idup will copy data. Where are the docs for that. Something else not in
the book - but then I guess there is a lot of detail not in the book.
Thank goodness for forums.

bob

On 21/09/2010 09:10, Pelle wrote:
> On 09/21/2010 09:29 AM, Bob Cowdery wrote:
>>   Hi
>>
>> I'm stuggling with immutable.
>>
>> I have a fixed size buffer which is used as a circular buffer of floats and is effectively double buffering data I wish to transfer to another thread. At an appropriate point I take the top half or bottom half of the buffer and send it to another thread.
>>
>> To do this I need to copy the data to an immutable transfer buffer to use in the send. I'm sure this is simple but I can't figure it.
>>
>> if I say something like:
>> float[] xfer = new float[512];
>> xfer = buffer[0 .. $/2];
>> tid.send(xfer);
>>
>> it rightly tells me 'thread local data not allowed'. If I make it:
>>
>> immutable (float)[] xfer;
>> xfer = buffer[0 .. $/2];
>> tid.send(xfer);
>>
>> it tells me 'can't implicitly convert float[] to immutable (float)[]'
>>
>> If I try a float by float copy into xfer it can't because I've said the buffer is immutable. In the first example I can't figure out how to convert the slice into an immutable copy which I think is what I should be doing.
>>
>> Can someone point me in the right direction.
>>
>> Thanks
>> Bob
>>
>
> immutable(float)[] xfer = buffer[0 .. $/2].idup;
> tid.send(xfer);
>
> Don't use assumeUnique for non-unique references.

Bob Cowdery <bob@bobcowdery.plus.com> wrote:

>  Ok, so assumeUnique when I read it a bit more will still share data but
> idup will copy data. Where are the docs for that. Something else not in
> the book - but then I guess there is a lot of detail not in the book.
> Thank goodness for forums.

The thing is, slices (as you did in xfer = buffer[0 .. $/2];) do not
copy data. I somehow misread your code as xfer[] = buffer[0 .. $/2];
which would copy data, if only xfer already had a length - sorry about that.

-- 
Simen