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digitalmars.D.learn - Immutable woes

reply Bob Cowdery <bob bobcowdery.plus.com> writes:
 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
Sep 21 2010
next sibling parent reply "Simen kjaeraas" <simen.kjaras gmail.com> writes:
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
Sep 21 2010
parent reply Bob Cowdery <bob bobcowdery.plus.com> writes:
 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.
Sep 21 2010
parent "Simen kjaeraas" <simen.kjaras gmail.com> writes:
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
Sep 21 2010
prev sibling parent reply Pelle <pelle.mansson gmail.com> writes:
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.
Sep 21 2010
parent reply Bob Cowdery <bob bobcowdery.plus.com> writes:
 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.
Sep 21 2010
parent "Simen kjaeraas" <simen.kjaras gmail.com> writes:
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
Sep 21 2010