Allegro利用Skill语言实现判断1个坐标是否在bBox构成的矩形框内的函数

/*****************获取叉乘*******************/
defun(GetCross (p1 p2 p)
	p1x=car(p1)
	p1y=nth(1 p1)
	p2x=car(p2)
	p2y=nth(1 p2)
	px=car(p)
	py=nth(1 p)
	let((res)
		res=(p2x-p1x)*(py-p1y)-(px-p1x)*(p2y-p1y)
	)	
)
/*****************判断1个坐标在bbox上*******************/
defun(IsInBbox (point bbox_list)
	line_line_spac=axlCNSGetSpacing("" "SIG3" 'line_line)
	line_line_spac=0
	startPoint=car(bbox_list)
	endPoint=nth(1 bbox_list)
	xstart=car(startPoint)-line_line_spac
	ystart=nth(1 startPoint)-line_line_spac
	xend=car(endPoint)+line_line_spac
	yend=nth(1 endPoint)+line_line_spac	
	let((res)
	res=GetCross(xstart:yend,xstart:ystart,point) * GetCross(xend:ystart,xend:yend,point) >= 0 && GetCross(xstart:ystart,xend:ystart,point) * GetCross(xend:yend,xstart:yend,point) >= 0;
	)
)
res=IsInBbox(100:100 list(0:0 200:200));x=t
res=IsInBbox(300:300 list(0:0 200:200));x=nil
res=IsInBbox(-300:-300 list(0:0 200:200));x=nil
axlUIWPrint(nil "res=%L" res)

 

var code = "9e2e90e7-f1f6-4793-a0dc-72d4b79bc121"
posted @ 2019-07-31 14:03  黑马Amos  阅读(31)  评论(0编辑  收藏  举报