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Windows环境下32位汇编语言程序设计,十分全面
Windows环境下32位汇编语言程序设计,十分全面-Windows Environment 32-bit assembly language programming, a very comprehensive
- 2022-11-03 12:05:03下载
- 积分:1
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在com中使用结构体枚举
在com中使用结构体枚举-used in the structure of Enumerate
- 2022-05-07 09:53:45下载
- 积分:1
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可以方便的进行三次以内方幂求和公式的自动推导 此代码为独创...
可以方便的进行三次以内方幂求和公式的自动推导 此代码为独创-Can easily be a power of less than three times the automatic derivation of the summation formula for the original code
- 2022-08-22 03:01:03下载
- 积分:1
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Visual C++ MFC Class Card Games book s source code, including flight chess, mahj...
Visual C++ MFC棋牌类游戏 书籍的源代码包括飞行棋,麻将等-Visual C++ MFC Class Card Games book s source code, including flight chess, mahjong,
- 2022-07-09 10:30:47下载
- 积分:1
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ssd4exercise5的作业答案,欢迎你的下载,谢谢了
ssd4exercise5的作业答案,欢迎你的下载,谢谢了-Operating ssd4exercise5 the answer, you are welcome to download, I would like to thank the
- 2023-03-18 05:15:03下载
- 积分:1
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NURBS曲线插补
本文较为深入地研究了 NURBS 曲线及其性质、相关参数计算方法。提出NURBS 曲线的阿当姆斯微分方程插补算法,并利用前、后差分结合代替微分
的方法,对初始迭代式改进得到参数快速递推式。利用参数快速递推式,同时辅以预估—校正法,通过适当的插补预处理与合理的近似计算,综合考虑轮廓
误差控制和进给加速度控制,实现了动态变步长的自适应 NURBS 曲线插补方式。可使机床在满足误差要求的前提下,以最大化进给速度对零件进行插补加
工,实现了插补精度和插补速度的优化,提高了加工质量和效率。在 NURBS插补加减速控制方面,本文利用 NURBS 曲线的对称性进行实时插补减速点的
预测,提出了插补前抛物线--直线--抛物线 S 型加减速控制新方法。其具有位置精度高、速度过渡平滑等优点,并且改变了原有控制方法的以弦线代替弧
线、每插补一步都要进行繁琐的曲线弧长计算的情况,有利于提高插补精度和实时性。
- 2022-03-20 21:20:11下载
- 积分:1
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An example of using URBI via LibURBI in C
An example of using URBI via LibURBI in C-An example of using URBI via LibURBI in C++
- 2022-02-15 01:49:10下载
- 积分:1
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matlab code for extracting grid
matlab code for extracting grid-based color moment and wavelet texture features. simple but effective.
- 2022-02-25 14:54:05下载
- 积分:1
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约瑟夫环(Joseph)问题的一种描述是:编号为1,2,…,n的n个人按顺时针方向围坐一圈,每人持有一个密码(正整数)。一开始人选一个正整数作为报数上限值m,从...
约瑟夫环(Joseph)问题的一种描述是:编号为1,2,…,n的n个人按顺时针方向围坐一圈,每人持有一个密码(正整数)。一开始人选一个正整数作为报数上限值m,从第一个人开始按顺时针方向自1开始顺序报数,报到m时停止报数。报m的人出列,将他的密码作为新的m值,从它在顺时针方向的下一个人开始重新从1报数,如此下去,直至所有人全部出列为止。试设计一个程序求出出列顺序-Joseph Central (Joseph) a description of the problem is: numbered 1,2, ..., n of n individuals according to a clockwise direction around a circle, each holding a password (positive integer). The beginning of a positive integer as a candidate at a number of upper limit of m, from the first individual to embark on the clockwise direction since the beginning of the order of 1 is reported the number of stops required to report back at a few m. Those who reported m out of his password as the new value of m, in a clockwise direction from the next person to start off again from the 1, and so on, until all that out until all. Try to design a procedure to derive the column order
- 2022-02-10 01:58:28下载
- 积分:1
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u_reg H”。
u_reg.h"
void Delay_MS(unsigned int ms)
void main()
{
asm(" STM #0000h,CLKMD ")
while(*CLKMD & 0x01 )
asm(" STM #40C7h,CLKMD ") //设置CPU运行频率=100M
/* 40C7h:5*clkin =100M
30c7h:4*clkin =80M
20c7h:3*clkin =60M
10C7h:2*clkin =40M
*/
asm(" stm #4240h, SWWSR ")
//2 wait except for on-chip program 1
asm(" stm #00a0h, PMST ") //MP/MC = 0, IPTR = 001,ovly=0
asm(" stm #0802h, BSCR ")
asm(" STM #0h,IMR ")
while(1)
{
Delay_MS(500)
asm(" RSBX XF ")
Delay_MS(500)
asm(" SSBX XF ")
}
-sdqwd
- 2022-02-04 22:56:44下载
- 积分:1