On Thu, Oct 11, 2012 at 04:30:42PM +0200, Richard Biener wrote:
> > The comment wants to say that, but doesn't do it correctly:
> > Operand 0 is a vector; the first element in the vector has the result.
> > Operand 1 is a vector. */
> > because obviously it doesn't have two operands, just one. So it should be
> > perhaps
> > Operand 0 is a vector.
> > The expression returns a vector of the same type, with the first
> > element in the vector holding the result of the reduction. */
> > ?
>
> Yes. It also should specify that the other elements are zero (or is that
> just your choice of "arbitrary"?)
That was my choice of arbitrary. Guess what the hw insns leave in there is
pretty much random, could be copies of the result in all elements, or
temporaries from pairwise reductions, etc.
Jakub
