vector expression question.

[Steve Kargl]

On Thu, Nov 17, 2011 at 11:30:52AM -0800, Mike Kumbera wrote:
> program test
> 
> integer, parameter :: n = 10
> integer :: i(n) = (/ 1, 2, 3, 4, 5, 5, 4, 3, 2, 1/)
> integer x(5),xser(5)
> 
> ! cumaddin
> xser = 0;
> do j = 1, n
>     xser (i (j)) =  xser (i (j)) + 1
> enddo
> write (*,*) xser
> 
> x = 0; 
> x(i) = x(i) + 1   ! same as the loop above

The RHS is evaulated before the LHS is assigned a value. 
So, what you have is essentially x(i) = (/0,0,0,0,0,0,0,0,0,0,0/) + 1.

-- 
Steve