值相同但不相同？
好的，我发现了一些最奇怪的错误。背景：
我试图找出两个矩形是否相交，如果它们在超过一个给定点上相交（即角点仅接触），它将把它添加到矩形相交列表中。
(if (and(and (<= (min ax1 ax2) bx1) (>= (max ax1 ax2) bx1))(and (<= (min ay1 ay2) by1) (>= (max ay1 ay2) by1)))(setq int 1))
(if (and(and (<= (min ax1 ax2) bx1) (>= (max ax1 ax2) bx1))(and (<= (min ay1 ay2) by2) (>= (max ay1 ay2) by2)))(setq int 1))
(if (and(and (<= (min ax1 ax2) bx2) (>= (max ax1 ax2) bx2))(and (<= (min ay1 ay2) by1) (>= (max ay1 ay2) by1)))(setq int 1))
(if (and(and (<= (min ax1 ax2) bx2) (>= (max ax1 ax2) bx2))(and (<= (min ay1 ay2) by2) (>= (max ay1 ay2) by2)))(setq int 1))
是的，我知道我应该使用cond，但cond不允许我需要超过1张支票。
好的，现在是设置和错误。
设置：
(setq acomp (nth 12 wlist))
(setq aname (nth 0 acomp))
(setq ax1 (nth 1 acomp))
(setq ay1 (nth 2 acomp))
(setq ax2 (nth 3 acomp))
(setq ay2 (nth 4 acomp))
(setq adir (nth 7 acomp))
(setq bcomp (nth 9 wlist))
(setq bname (nth 0 bcomp))
(setq bx1 (nth 1 bcomp))
(setq by1 (nth 2 bcomp))
(setq bx2 (nth 3 bcomp))
(setq by2 (nth 4 bcomp))
(setq bdir (nth 7 bcomp))
("P3" 202.0 175.0 226.0 178.0 202.0 176.5 "X")
"P3"
202.0
175.0
226.0
178.0
"X"
("W10" 198.0 142.0 202.0 254.0 200.0 142.0 "Y")
"W10"
198.0
142.0
202.0
254.0
"Y"错误：
_$ ax1
202.0
_$ ax2
226.0
_$ bx2
202.0
_$ (min ax1 ax2)
202.0
_$ (<= (min ax1 ax2) bx2)
nil
您最好的选择可能是在计算中引入“模糊”因子，您可以使用“相等”函数来实现。
如果您希望清理代码，请尝试包装所有（和…）函数转换为单个（和…）函数或（或…）。这样，每个人都将被评估，而不需要有一堆if。这是一个很好的提示；如果你有很多非常相似的if，考虑把它们变成某种循环；你也许可以节省一些空间并将其清理干净。
希望这有帮助，至少有一点！ 方程得出：
Command: (= pi 3.1415926535897932)
T
Command: (= pi 3.1415926535897933)
T
但这再次带来了错误的矩形相交。
谢谢你教我一个新命令。
有没有办法把四舍五入到最接近的1/64英寸或1/32英寸。这将通过将交点转换为预期值来停止假矩形交点。
此外，为了清晰起见，这里有一些修订的代码。更好地理解我在做什么。
_$ (Equal (min ax1 ax2) bx2 0.00000001)
T
Ok全部固定
谢谢你的帮助。
圆形函数：
(setq int 0)
(if
(and
(and (<= (min ax1 ax2) bx1)
(>= (max ax1 ax2) bx1)
);;point 1's x is between or on Rectangle
(and (<= (min ay1 ay2) by1)
(>= (max ay1 ay2) by1)
);;point 1's y is between or on Rectangle
);;point 1 is between or on Rectangle
(setq int (+ int 1));; Rectangles intersect add to intesections
);;end if
(if
(and
(and (<= (min ax1 ax2) bx1)
(>= (max ax1 ax2) bx1)
);;point 2's x is between or on Rectangle
(and (<= (min ay1 ay2) by2)
(>= (max ay1 ay2) by2)
);;point 2's y is between or on Rectangle
);;point 2 is in or on Rectangle
(setq int (+ int 1));; Rectangles intersect add to intesections
);;end if
(if
(and
(and (<= (min ax1 ax2) bx2)
(>= (max ax1 ax2) bx2)
);;point 3's x is between or on Rectangle
(and (< (min ay1 ay2) by1)
(> (max ay1 ay2) by1)
);;point 3's y is between or on Rectangle
);;point 3 is in or on Rectangle
(setq int (+ int 1));; Rectangles intersect add to intesections
);;end if
(if
(and
(and (<= (min ax1 ax2) bx2)
(>= (max ax1 ax2) bx2)
);;point 4's x is between or on Rectangle
(and (<= (min ay1 ay2) by2)
(>= (max ay1 ay2) by2)
);;point 4's y is between or on Rectangle
);;point 4 is in or on Rectangle
(setq int (+ int 1));; Rectangles intersect add to intesections
);;end if
矩形检查：
(defun round (number xth / ro x y out)
(if (> xth 0)
(progn
(setq ro (rem number 1))
(setq x (- number ro))
(if (>= (rem (* (expt 10 xth) ro) 1) 0.5) (setq add 1) (setq add 0))
(setq y (/(- (* (expt 10 xth) ro) (- (rem (* (expt 10 xth) ro) 1) add)) (expt 10 xth)))
(setq out (+ x y))
)
)
(if (= xth 0)
(progn
(setq ro (rem number 1))
(setq out (- number ro))
)
)
(if (< xth 0)
(progn
(setq xth (* xth -1))
(setq ro (rem (/ number (expt 10 xth)) 1))
(setq out (* (- (/ number (expt 10 xth)) ro)(expt 10 xth)))
)
)
out
)
输入：
(setq int 0)
(if(and(and(<= (min ax1 ax2) bx1)(>= (max ax1 ax2) bx1))(and (<= (min ay1 ay2) by1)(>= (max ay1 ay2) by1)))(setq int (+ int 1)))
(if(and(and(<= (min ax1 ax2) bx1)(>= (max ax1 ax2) bx1))(and (<= (min ay1 ay2) by2)(>= (max ay1 ay2) by2)))(setq int (+ int 1)))
(if(and(and(<= (min ax1 ax2) bx2)(>= (max ax1 ax2) bx2))(and (<= (min ay1 ay2) by1)(>= (max ay1 ay2) by1)))(setq int (+ int 1)))
(if(and(and(<= (min ax1 ax2) bx2)(>= (max ax1 ax2) bx2))(and (<= (min ay1 ay2) by2)(>= (max ay1 ay2) by2)))(setq int (+ int 1)))
(if (> int 1) (progn))
输出：
_$ (setq acomp (nth 1 wlist))
(setq aname (nth 0 acomp))
(setq ax1 (round (nth 1 acomp) 4))
(setq ay1 (round (nth 2 acomp) 4))
(setq ax2 (round (nth 3 acomp) 4))
(setq ay2 (round (nth 4 acomp) 4))
(setq adir (nth 7 acomp))
(setq bcomp (nth 0 wlist))
(setq bname (nth 0 bcomp))
(setq bx1 (round (nth 1 bcomp) 4))
(setq by1 (round (nth 2 bcomp) 4))
(setq bx2 (round (nth 3 bcomp) 4))
(setq by2 (round (nth 4 bcomp) 4))
(setq bdir (nth 7 bcomp))
("W2" 326.0 6.0 330.0 138.0 328.0 6.0 "Y")
"W2"
326.0
6.0
330.0
138.0
"Y"
("W1" 22.0 6.0 326.0 10.0 22.0 8.0 "X")
"W1"
22.0
6.0
326.0
10.0
"X"
is\u point\u in\u box测试可以压缩很多
0
nil
nil
1
2
-大卫 “圆形”函数也可以极大地压缩：
;;;ARG -> Test_pt Corner_Pt1 Corner_Pt2
;;;RET -> T nil
(defun is_pt_in_box (tp p1 p2 / x1 x2 y1 y2)
(setq x1 (car p1)x2 (car p2)
y1 (cadr p1) y2 (cadr p2))
(and (>= (max x1 x2) (car tp) (min x1 x2))
(>= (max y1 y2) (cadr tp) (min y1 y2))))
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