Hello, In your example the condition is necessary: if N is some large prime number, you cannot create grid/block pair which contains exactly N total threads; so you have to skip excessive ones somehow. Moreover, the "if" statement is not expensive by itself, it becomes expensive if it causes execution to diverge inside a single warp. Here it will diverge execution in at most one warp (and not for much), so will not affect the performance significantly. Maybe it will be even optimized out completely by branch predication, but I am not sure about that.
Best regards, Bogdan On Wed, Sep 28, 2011 at 1:08 AM, ericyosho <ericyo...@gmail.com> wrote: > I'm not sure if it is the right place, but since it is so elementary, > I just appreciate some explanation. > So in every CUDA tutorial example, e.g., to double each element in an > array, in kernel function, we have the following lines: > > int idx = // calculate a unique value for each thread > if (idx < N) // N is the number of elements of an array > a[idx] *= 2; > > "if branch" is a rather expensive operation, why do we want each > thread to go for this check? > Since on each device, only one kernel function is allowed to evaluate > at a time, why don't we let each thread double its own associated > value, and afterwards we simply copy N elements back to the host. > Basically, we just omit the "if" check, and go for the "double values" > line unconditionally. > > It seems this approach is more straightforward. > Do I miss anything? > > Best, > Zhe Yao > -------------- > Department of Electrical and Computer Engineering > McGill University > Montreal, QC, Canada > H3A 2A7 > > zhe....@mail.mcgill.ca > > _______________________________________________ > PyCUDA mailing list > PyCUDA@tiker.net > http://lists.tiker.net/listinfo/pycuda > _______________________________________________ PyCUDA mailing list PyCUDA@tiker.net http://lists.tiker.net/listinfo/pycuda
