之前写文章想插代码时，各种不爽，都想换地方写了，然后发现原来这里支持markdown，所以入坑了，相关介绍如下：
简书markdown
简介
Cohen–Sutherland是一个线段裁剪算法
原理
将窗口区域分为9个部分，每个部分给一个区域码，然后计算线段两端端点的区域码，根据区域码来选择抛弃线段
Paste_Image.png
- 两端点都在视口区域内，区域码相或为0，接受
- 两端点至少共享一个不可见区域，区域码相与不为1，拒绝
实现
typedef int OutCode;
const int INSIDE = 0; // 0000
const int LEFT = 1; // 0001
const int RIGHT = 2; // 0010
const int BOTTOM = 4; // 0100
const int TOP = 8; // 1000
// Compute the bit code for a point (x, y) using the clip rectangle
// bounded diagonally by (xmin, ymin), and (xmax, ymax)
// ASSUME THAT xmax, xmin, ymax and ymin are global constants.
//判断端点的区域码
OutCode ComputeOutCode(double x, double y)
{
OutCode code;
code = INSIDE; // initialised as being inside of clip window
if (x < xmin) // to the left of clip window
code |= LEFT;
else if (x > xmax) // to the right of clip window
code |= RIGHT;
if (y < ymin) // below the clip window
code |= BOTTOM;
else if (y > ymax) // above the clip window
code |= TOP;
return code;
}
// Cohen–Sutherland clipping algorithm clips a line from
// P0 = (x0, y0) to P1 = (x1, y1) against a rectangle with
// diagonal from (xmin, ymin) to (xmax, ymax).
void CohenSutherlandLineClipAndDraw(double x0, double y0, double x1, double y1)
{
// compute outcodes for P0, P1, and whatever point lies outside the clip rectangle
OutCode outcode0 = ComputeOutCode(x0, y0);
OutCode outcode1 = ComputeOutCode(x1, y1);
bool accept = false;
while (true) {
if (!(outcode0 | outcode1)) { //相或为0，接受并且退出循环
accept = true;
break;
} else if (outcode0 & outcode1) { // 相与为1，拒绝且退出循环
break;
} else {
// failed both tests, so calculate the line segment to clip
// from an outside point to an intersection with clip edge
double x, y;
//找出在界外的点
OutCode outcodeOut = outcode0 ? outcode0 : outcode1;
// 找出和边界相交的点
// 使用点斜式 y = y0 + slope * (x - x0), x = x0 + (1 / slope) * (y - y0)
if (outcodeOut & TOP) { // point is above the clip rectangle
x = x0 + (x1 - x0) * (ymax - y0) / (y1 - y0);
y = ymax;
} else if (outcodeOut & BOTTOM) { // point is below the clip rectangle
x = x0 + (x1 - x0) * (ymin - y0) / (y1 - y0);
y = ymin;
} else if (outcodeOut & RIGHT) { // point is to the right of clip rectangle
y = y0 + (y1 - y0) * (xmax - x0) / (x1 - x0);
x = xmax;
} else if (outcodeOut & LEFT) { // point is to the left of clip rectangle
y = y0 + (y1 - y0) * (xmin - x0) / (x1 - x0);
x = xmin;
}
// Now we move outside point to intersection point to clip
// 为什么继续循环，两个端点都有可能在外面
if (outcodeOut == outcode0) {
x0 = x;
y0 = y;
outcode0 = ComputeOutCode(x0, y0);
} else {
x1 = x;
y1 = y;
outcode1 = ComputeOutCode(x1, y1);
}
}
}
if (accept) {
// Following functions are left for implementation by user based on
// their platform (OpenGL/graphics.h etc.)
DrawRectangle(xmin, ymin, xmax, ymax);
LineSegment(x0, y0, x1, y1);
}
}
其他
在计算交点的时候，之前有这样的做法：求两点的中点，然后用中点和另一个点的位运算判断，然后再用中点和另一个点位运算判断，一直这样下去，直到中点区域交点。这样主要是效率考虑：以前cpu做位运算速度高于乘法除法等，乘法就是加法的累积，所以求中点（右移）比点斜式的计算快。
我在VS上做了测试，却发现加减乘除的速度都几乎一致
#include "stdafx.h"
#include <iostream>
#include <time.h>
int _tmain(int argc, _TCHAR* argv[])
{
int wait;
int times = 100000000;
float res = 0;
clock_t nowtime, endtime;
nowtime = clock();
while(times--)
{
res = 201+311;//这里改成加减乘除结构几乎一致
}
endtime = clock();
std::cout<<"the cal is spend " << (endtime - nowtime)<< " mili seconds" << nowtime<<std::endl<<endtime;
std::cin>>wait;
return 0;
}
网上查找相关资料后，应该是现在的cpu已经有乘法器等相关硬件支持，而且编辑器也足够先进，所以加减乘除速度几乎没什么区别了
