代码搜索:while
找到约 10,000 项符合「while」的源代码
代码结果 10,000
www.eeworm.com/read/476373/6758289
c jac.c
/* Author: Joseph M. Reagle Jr.
* Purpose: To use the Jacobi function to find the probably prime numbers
* within a certain numberic range. The data types can prove to be problematic
* for large
www.eeworm.com/read/476373/6758290
c knap.c
/* Author: Joseph M. Reagle Jr.
* Purpose: To exhaustively break knapsack ciphers.
*/
/* The following code is provided as is.
* It may be copied and modified freely provided that
*Joseph M
www.eeworm.com/read/476444/6762556
c ircom.c
/* 晶振:11.0569MHz */
#include
#define uchar unsigned char
uchar data IRcode[4]; //定义一个4字节的数组用来存储代码
uchar CodeTemp; //编码字节缓存变量
uchar i,j,k; //延时用的循环
www.eeworm.com/read/476128/6763320
c main.c
#define uchar unsigned char //定义一下方便使用
#define uint unsigned int
#define ulong unsigned long
#include //包括一个52标准内核的头文件
sbit P10 = P1^0; //头文件中没有定义的IO就要自己来定义了
sbit P11 = P1^1;
sb
www.eeworm.com/read/476128/6763554
c main.c
#define uchar unsigned char //定义一下方便使用
#define uint unsigned int
#define ulong unsigned long
#include //包括一个52标准内核的头文件
char code dx516[3] _at_ 0x003b;//这是为了仿真设置的
sbit P10=P1^
www.eeworm.com/read/476156/6763649
m router.m
%配合floyd算法的后续程序,s为源点,t为宿点
%L为长度,R为路由
function [L,R]=router(D,path,s,t)
L=zeros(0,0);
R=s;
while 1
if s==t
L=fliplr(L);
L=[0,L];
return
end
L=[L,D(s,t)];
www.eeworm.com/read/476277/6764908
txt kmp字符串匹配.txt
#include
#include
int next[20];
int main(){
char a[20]="abcdefghijk";
char b[20]="abcdefgg";
int i,j;
i=0;next[0]=0;j=0;
while(i
www.eeworm.com/read/265088/11283881
m chap7_2a.m
%BP Training for MIMO and Multi-samples
clear all;
close all;
xite=0.50;
alfa=0.05;
w2=rands(6,2);
w2_1=w2;w2_2=w2_1;
w1=rands(3,6);
w1_1=w1;w1_2=w1;
dw1=0*w1;
I=[0,0,0,0,0,0]';
Iou
www.eeworm.com/read/265015/11285989
c 单词排序.c
#include
#include
main()
{
FILE *in,*out;
char c,s[100][50]={0},a[100]={0};
int i=0,j=0,k;
in=fopen("sort.in","r");
out=fopen("sort.out","w");
c=fgetc(in);
while(
www.eeworm.com/read/265015/11285996
c 打印文件末尾n行.c
#include
#include
main(int argc,char *argv[])
{
FILE *in;
char s[1000][10000]={0},c;
int j=0,i=0,n=0;
if(argc!=3||argv[1][0]!='-'||(in=fopen(argv[2],"r"))==NULL