功能描述
功能很简单，程序更简单：
简单的蜂鸣器实验程序：本程序通过在P1.1输出一个音频范围的方波，驱动实验板上的蜂鸣器发出蜂鸣声，其中包含有一个定时模块，作用是使输出的方波频率在人耳朵听觉能力之内的20KHZ以下，如果没有这个延时程序的话，输出的频率将大大超出人耳朵的听觉能力，我们将不能听到声音。更改定时常数，可以改变输出频率，也就可以调整蜂鸣器的音调。
Archive for ‘边走编程’ Category
April 29, 2009
功能描述
由p1口外接按键开关，p3.3接蜂鸣器组成。
1 2 3 4 5 6 7 8 9 | org 8000h ;定位 ljmp start org 8100h start: mov sp, #60h mov tmod, #10h ;置T1为方式1 ; mov tl1, #010h ; mov th1, #010h ;设初值，定时为50ms setb tr1 ;启动定时器T1 |
功能描述
正常情况下(p3.2为高电平), p3.3的电平以一定的频率连续翻转，当p3.2为低电平时，触发int0外部中断，进入中断服务程序，另p3.3保持低电平，并将p1端口加1。p3.2外接一个按键开关时，中断服务程序有防抖动的功能，这时通过一定的延时(通常为10ms–20ms)来实现的。
程序
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 | org 8000h ;定位 ljmp start org 8003h ;中断向量 ljmp int_0 org 8100h start: mov sp, #60h setb ex0 ;开int0中断 setb ea ;开总中断 mov tcon, #00h ;低电平触发 mov r3, #00h mov a, r3 cpl a loop3: ;死循环 mov p1, a cpl p3.3 ;翻转 lcall DELAY ;延时 sjmp loop3 int_0: ;中断服务子程序 push psw lcall DELAY ;防抖动 setb p3.3 inc r3 mov a, r3 cpl a mov p1, a jnb p3.2, $ ;查询，防止多次中断 lcall DELAY ;还是防抖动 pop psw reti DELAY: PUSH 06H PUSH 07H MOV R7, #70H LABEL1: MOV R6, #70H LABEL2: DJNZ R6, $ DJNZ R7, LABEL1 POP 07H POP 06H RET END |
89C51单片机，每计时一秒将P1端口翻转一次，晶振频率12MHz，中断方式实现。这里另外一个查询方式的实现
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 | org 8000h ;定位 ljmp start org 801bh ;中断向量 ljmp int_t1 org 8100h start: mov sp, #60h mov tmod, #10h ;置T1为方式1 mov tl1, 0fch mov th1, #4bh ;设初值，定时为50ms mov r1, #20 ;中断次数，以使定时为1ms setb tr1 ;启动定时器T1 clr a ;累加器清零 setb et1 ;T1中断允许 setb ea ;开总外部中断标志 sjmp $ ;等待中断 int_t1: ;中断服务子程序 push psw ;保存程序状态字 mov tl1, #0fch ;重装计数器 mov th1, #4bh djnz r1, exit ;1s时间未到，返回主程序，等待下一次中断 mov r1, #20 ;1s时间到，重装r1 cpl a ;a取反 mov p1, a ;输出a exit: pop psw ;程序状态字出栈 reti DELAY: PUSH 06H PUSH 07H MOV R7, #70H LABEL1: MOV R6, #70H LABEL2: DJNZ R6, $ DJNZ R7, LABEL1 POP 07H POP 06H RET END |
89C51单片机，每计时一秒将P1端口翻转一次，晶振频率12MHz，这里另外一个中断方式的实现
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | org 8000h ;定位 ljmp start org 8100h start: mov sp, #60h mov r1, #20 clr a mov tmod, #10h ;置T1为方式1 loop: mov tl1, 0fch mov th1, #4bh ;设初值，定时为50ms setb tr1 ;启动定时器T1 jnb tf1, $ ;查询TF1是否溢出 clr tf1 ;清除溢出标志 djnz r1, loop ;1s时间未到，重装计数器 mov r1, #20 ;1s时间到，重装r1 cpl a ;a取反 mov p1, a ;输出a ljmp loop END |
April 28, 2009
POJ 的一道题，算法很简单：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 | #include <string> #include <iostream> using namespace std; int main() { int n; char b2h[17] = "0123456789ABCDEF"; int flag[4] = {1,2,4,8}; cin>>n; while (n--) { string s;cin>>s; int len = s.length(); int mod = len%4; int num = 0; for (int i=mod;i>0;i--) { if (s[mod-i]=='1') { num += flag[i-1]; } } if (mod) cout<<b2h[num]; for (int i=mod;i<len;i+=4) { num = 0; for (int j=0;j<4;j++) { if (s[i+j]=='1') { num += flag[3-j]; } } cout<<b2h[num]; } cout<<endl; } return 0; } |
April 27, 2009
运行期(runtime)判断
下面的程序可以在运行期判断 endianess:
int IsBigEndian (void) { static const int v =1; return *(char*)&v?0:1; } |
Description
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,…,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
April 25, 2009
Introduction
A good way to get into an argument with a computer programmer is to attempt to explain why the language they are using is not as good as the one you are using. Most of the programmers I know are positively religious over their Operating System (see my other article), their development language and finally their text editor.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 | bool fun(unsigned n) { assert(n >0); //乘方不为正 if (!(n & (n-1))) //n的二进制只有一个1，即n==2^m; { int offset = 0;//"1"位的偏移量 while (n != 1) { n>>=1; offset++; } if (offset && !(offset&1))//偏移量为偶数，即 n == 2^(2*m),亦即n == 4^m; { return true; } } return false; } |