Re: Distance from point to box
От | Alexander Korotkov |
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Тема | Re: Distance from point to box |
Дата | |
Msg-id | CAPpHfdsDuvucZGwvtdOUCQ8x1crmLhaRMEiHDKVJiUx38VDE_Q@mail.gmail.com обсуждение исходный текст |
Ответ на | Re: Distance from point to box (Fabien COELHO <coelho@cri.ensmp.fr>) |
Ответы |
Re: Distance from point to box
|
Список | pgsql-hackers |
On Wed, Jul 30, 2014 at 4:06 PM, Fabien COELHO <coelho@cri.ensmp.fr> wrote:
ISTM that you miss the projection on the segment if dx=0 or dy=0.double dx = 0.0, dy = 0.0;
if (point->x < box->low.x)
dx = box->low.x - point->x;
if (point->x > box->high.x)
dx = point->x - box->high.x;
if (point->y < box->low.y)
dy = box->low.y - point->y;
if (point->y > box->high.y)
dy = point->y - box->high.y;
return HYPOT(dx, dy);
I feel myself quite tangled.
Could anybody clarify it for me? Did I miss something? Thanks.
I don't need to find projection itself, I need only distance. When dx = 0 then nearest point is on horizontal line of box, so distance to it is dy. Same when dy = 0. When both of them are 0 then point is in the box.
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With best regards,
Alexander Korotkov.
With best regards,
Alexander Korotkov.
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