On Sat, 11 Feb 2012 17:18:21 +0100, Ali Çehreli <[email protected]> wrote:
On 02/11/2012 12:56 AM, Martin Nowak wrote:
> On Sat, 11 Feb 2012 02:31:29 +0100, Ali Çehreli <[email protected]>
wrote:
>
>> Sorry for the double-post; I have asked the same question on D.learn
>> earlier but I think this is more of a question to this forum.
>>
>> Tested on Ubuntu 11.10 64-bit dmd.
>>
>> The following program gets stuck during the writeln() call.
>>
>> - Note that the foo() call alone works fine.
>>
>> - Also note that the program works fine when there is no writeln()
>> call nor foo() call. All elements get processed in that case and the
>> results are ignored.
>>
>> Am I using taskPool.map incorrectly or is this a bug? Can you help
>> identify where the problem may be? How is writeln() using the range
>> differently than foo() to cause this behavior?
>>
>> import std.stdio;
>> import std.parallelism;
>> import core.thread;
>>
>> int func(int i)
>> {
>> writeln("processing ", i);
>> return i;
>> }
>>
>> void main()
>> {
>> auto results = taskPool.map!func([1,2,3,4,5,6,7,8], 2);
>>
>> writeln(results); // <-- Gets stuck HERE
>>
>> foo(results); // this works fine
>> }
>>
>> void foo(R)(R range)
>> {
>> for ( ; !range.empty; range.popFront()) {
>> writeln(range.front);
>> }
>> }
>>
>> Thank you,
>> Ali
>
> Yeah, you have a deadlock in there, it's somewhat hidden though.
> The issue is that writeln will take lock on stdout once. This will
deadlock
> with the lazy processing of the map range.
Thank you.
I was trying to visualize the semi-lazy nature of taskPool.map. Now I
get what I want when the writeln() call in main() is changed to be on
stderr:
import std.stdio;
import std.parallelism;
import core.thread;
int func(int i)
{
writeln("processing ", i);
Thread.sleep(dur!"seconds"(1));
return i;
}
void main()
{
auto results = taskPool.map!func([1,2,3,4,5,6,7,8], 2);
stderr.writeln(results); // <-- now on stderr
}
Now the output hints at how taskPool.map is semi-lazy:
processing 1
processing 2
[processing 3
1, 2processing 4
, 3processing 5
, 4processing 6
, 5, 6processing 7
processing 8
, 7, 8]
Good. :)
Ali
Going with foreach(e; results) will work too and you can write to the
right output.
