Ubuntu 11.10 64-bit dmd.

The following program gets stuck during the writeln() call.

- The foo() call alone works fine.

- 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?

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

    foo(results); // this works fine
}

void foo(R)(R range)
{
    for ( ; !range.empty; range.popFront()) {
        writeln(range.front);
    }
}

Thank you,
Ali

On 02/10/2012 10:28 AM, Ali Çehreli wrote:
> Ubuntu 11.10 64-bit dmd.
>
> The following program gets stuck during the writeln() call.
>
> - The foo() call alone works fine.
>
> - 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?
>
> 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
>
> foo(results); // this works fine
> }
>
> void foo(R)(R range)
> {
> for ( ; !range.empty; range.popFront()) {
> writeln(range.front);
> }
> }
>
> Thank you,
> Ali

I have asked the same question on the main D newsgroup. It turns out, since the writeln() calls in func() and main() share the same resource (namely stdout), the main thread and the task threads get dead-locked on an output synchronization lock.

Changing the call in main to be on stderr fixes the issue and I get what I want. Now I can see how taskPool.map works semi-lazily:

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
}

The output:

processing 1
processing 2
[1, 2processing 3
processing 4
, 3, 4processing 5
processing 6
, 5, 6processing 7
processing 8
, 7, 8]

std.parallelism is only for when the tasks are truly independent from each other. I had forgotten that using stdout would violate that condition.

Ali