博文
将整数转化为一个字符串(2007-12-16 18:13:00)
摘要：#include<stdio.h> #define LENGTH 10 void IntToChar(int *p,int k,int n) { if(k!=0) { *(p+k--)=n%10+48;/*在ascii中的码的顺序,相差的值为48*/ IntToChar(p,k,n/10); } else { *(p+k)=n+48; return; } /*如果n是一位数，则*(p+0),即a[0]等于此数,对于多位数,已除到数的最高位*/ } void main() { int a[LENGTH],i,j=0,n,m; /*j用于判断输入数n是几位数*/ printf("Please input the integer wanted to change:\n"); scanf("%d",&n); m=n; /*m仅为方便下步while循环中判断n为几位数的参数*/ while(m/10!=0) /*while循环中判断n为几位数*/ { m=m/10; j++; } IntToChar(a,j,n); /*a[LENGTH]为存储转化后字符的数组*/ printf("The changed char is:\n"); for(i=0;i<=j;i++) printf("%c",a[i]); }
/*说明：*//*由于各种c编译器对int型数据的取值范围不一样,可能会有不同的结果*/ /*Turbo C 中int的范围为不能超过32767*/ ......
关于冒泡法(2007-11-27 11:11:00)
摘要：看了文涛的blog,感触很深,觉得程序的有些细节问题可以自己通过编程来了解,也更细致地了解它的执行过程,同时也当作是一种测试,下面是一个排序的测试程序
#include <stdio.h>void main(){ int a[11]; /* 第0号元素不用 */ int i, j,k, cup; printf("Input 10 numbers:\n");
for(i = 1; i < 11; i++) scanf("%d", &a[i]); for(i = 1; i <= 9; i++) /* 趟数 */ { printf("i=%d\n\n",i); for(j = 1; j <= 10 - i; j++) /* 每趟要比较数 */ { if(a[j] > a[j+1]) /* 前面数大于后面数刚对调 */ { cup = a[j]; a[j] = a[j+1]; &nbs......
vhdl实现的多循环方式彩灯(2007-11-20 15:29:00)
摘要：------the top file of cyc_led designlibrary ieee;use ieee.std_logic_1164.all;use ieee.std_logic_unsigned.all;use ieee.std_logic_arith.all;entity cyc_led isport(contr : in std_logic_vector(1 downto 0); clk : in std_logic; light : out std_logic_vector(7 downto 0));end entity cyc_led;
architecture cyc_light of cyc_led is component mode1 is---mode1 port(clk1 : in std_logic; light1 : out std_logic_vector(7 downto 0)); end component; component mode2 is---mode2 port(clk2 : in std_logic; light2 : out std_logic_vector(7 downto 0)); end component; component mode3 is ---mode3 port(clk3 : in std_logic; light3 : out std_logic_vector(7 downto 0)); end component; component mode4 is ---mode4 port(clk4 : in std_logic; light4 : out std_logic_vector(7 downto 0)); end component; signal temp1,temp2,temp3,t......
VHDL的任意整数且占空比为50%分频代码(2007-11-19 17:35:00)
摘要：说明如下：
1.其中top file 为 division，其中的clk_com是比较的频率，用它来和分频后波形进行比较，便于观察，
2.any_enve为任意偶数分频文件
3.any_odd为任意奇数分频文件
4.是一个用于2进制与8进制的译码器，我用它来显示在数码管上当前到底是多少分频
5.以下代码在开发板上实验过，请大家放心使用，欢迎转载，但请注明出处，另外说明由于用的是quartus7.1编辑的，中间无法加中文注释，请大家慢慢读了；以下是代码：
------the top file of the design divisionlibrary ieee;use ieee.std_logic_1164.all;use ieee.std_logic_arith.all;use ieee.std_logic_unsigned.all;entity division is port (input : in std_logic_vector(7 downto 0); clk : in std_logic; clk_out : out std_logic; clk_com : out std_logic; led1: out std_logic_vector(6 downto 0); led2: out std_logic_vector(6 downto 0); led3: out std_logic_vector(6 downto 0));end entity division;--------------------------------------------------
architecture freq of division is component decoder is----decoder port(bin : in std_logic_vector(2 downto 0); de : out std_logic_vector(6 downto 0)); end component; component any_......
数码管动态显示(2007-11-07 12:45:00)
摘要：#include <reg51.h>typedef unsigned char byte;// 0-255 typedef unsigned char word;//0-65535static byte arry_display[5];byte table[10]={0x42,0xee,0x58,0x68,0xe4,0x61,0x41,0xea,0x40,0x60};//0-9//显示函数void display(void){byte position=0xfe;byte i,j,temp;for(i=0;i<4;i++)//4数码管轮留导通 { temp=arry_display[i]; temp=table[temp]; for(j=0;j<200;j++)//延时 { P2=position;P0=temp; } position<<=1; position|=0x01;//以保证循环点亮 }//position=0xfe;}///将10进制转化为BCD
int cov_bcd(unsigned int n){ arry_display[0]=n/1000;//千位 arry_display[1]=(n/100)%10;//百位 arry_display[2]=(n/10)%10;//十位 arry_display[3]=n%10; //个位}void main(void){unsigned int a;a=0x00; while(1) { for(;a<9999;a++) { cov_bcd(a); display(); //for(j=0;j<3000;j++); } a=0x00; }}......
vhdl实现的8位可逆计数器(2007-11-01 19:55:00)
摘要：能实现连续脉冲还手动脉冲计数，在实验板上实验过~！
程序很简单，相应的注释就没加了，
library ieee;use ieee.std_logic_1164.all;use ieee.std_logic_unsigned.all;entity counter is port(rst,clk,clk1: in std_logic; con : std_logic_vector(1 downto 0); output : out std_logic_vector(7 downto 0)); end entity counter;
architecture behave of counter is begin process(rst,con,clk,clk1) variable temp1 : std_logic_vector(7 downto 0); begin if rst = '1' then temp1 := (others => '0'); else case con is when "10" =>--手动顺记数 if clk'event and (clk='1') and (clk'last_value='0') then if temp1 < 255 then temp1 := temp1 + 1; else temp1 := (others => '0'); end if; ......
猴子吃桃问题代码(2007-10-26 20:32:00)
摘要：#include <stdio.h>int remain(int i){ int j; if(i==10) j=1; else if(i<10&&i>0) j=(2*remain(i+1))+2; else if (i<=0) j=remain(1); else if(i>10) j=0; return(j);}void main(){int a,b;printf("please input the day you want to kown how many peach remained:\n");scanf("%d",&a);if(a>10)printf("The peach is eaten out.\n");else if(a<=0)printf("The peach is not eatean,so have %d peach.\n",remain(1));else { for(b=a;b>=1;b--) printf("The %d day reamins %d peach.\n",b,remain(b)); }}......
第一个实验-8路开关控制8个灯(2007-10-26 20:22:00)
摘要：library ieee;use ieee.std_logic_1164.all;entity switch_led is port(key : in std_logic_vector(7 downto 0); light : out std_logic_vector(7 downto 0));end entity switch_led;
architecture behav of switch_led is begin process(key) begin for i in 7 downto 0 loop light(i) <= key(i);--the led is lighting when the votage is high; end loop; end process; end architecture behav;......
38译码器的几种描述方法(2007-10-26 20:21:00)
摘要：library ieee;use ieee.std_logic_1164.all;use ieee.std_logic_unsigned.all;entity decoder38 is port (Q0: out std_logic_vector(7 downto 0); Q1: in std_logic_vector(2 downto 0); en: in std_logic);end entity decoder38;-----the num.1 method to descriptarchitecture code1 of decoder38 is begin process(Q1,en) variable temp : std_logic_vector(7 downto 0); begin if en = '1' then temp := "ZZZZZZZZ";---"en" is effective on low voltage; else case Q1 is when "000" => temp := "00000001"; when "001" => temp := "00000010"; when "010" => temp := "00000100"; when "011" => temp := "00001000"; when "100" => temp := "00010000"; when "101" => temp := "00100000"; when "110" => temp := "01000000";&......
魔方阵的c语言算法(2007-09-26 16:41:00)
摘要：
魔方阵的c语言算法:#include "stdio.h"#define Q 100int a[Q][Q];FILE *fp;void swap(int *a,int *b)/*交换函数*/{ int t; t=*a; *a=*b; *b=t;}
void Magic2Kplus1(int t,int H,int L,int N){/* 阶数N是奇数魔方阵 其中int t,表示：从t开始向魔方阵中填数 int H,int L分别表示从从魔方阵中的哪个位置开始填数 int N 表示魔方阵的阶数下面举例子说明以3阶魔方阵为例第1步： 中间填 1x 1 xx x xx x x第2步：数2应向1的右上角填但是没地方了就移到右下角x 1 xx x xx x 2第3步：数3应向2的右上角填但是没地方了就移到2上一行的最左边x 1 x3 x xx x 2第4步：数4应向3的右上角填但是这个地方已经有数字1了就移到3下边x 1 x3 x x4 x 2第5步：数5向4的右上角填x 1 x3 5 x4 x 2第6步 数6向5的右上角填x 1 63 5 x4 x 2第7步：数7应向6的右上角填但是没地方了......