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/324231/13278247
bak test.bak
#include
#define uchar unsigned char
#define uint unsigned int
sbit P0_7=P0^7;
void delay(uint x);
void main(void)
{
uchar code tab1[6]={0xc0,0xf9,0xa4,0xb0,0x99,0x92};
uchar
www.eeworm.com/read/239388/13282877
cpp disk.cpp
///////////////////////////////////////////////////////////////////////////
// 题目相当于给定(1/2)*n(n-1)个元素PiPj的集合A和(1/2)*n(n-1)个 //
// 元素d(i,j)的集合B，要求从A、B集合中各选一个元素配对求积，使 //
// 这
www.eeworm.com/read/239373/13283226
pl primenum.pl
const true=1,false=0;
var x,y,m,n,pf;
procedure prime;
var i,f;
procedure mod;
x:=x-x/y*y;
begin f:=true;
i:=3;
while i
www.eeworm.com/read/239373/13283238
pl cock.pl
const z=0;
var head,foot,cock,rabbit,n;
begin
n:=z;
read(head,foot);
cock:=1;
while cock
www.eeworm.com/read/324106/13286903
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/324106/13287851
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/239252/13292005
c delay.c
#include "DELAY.H"
static void delay100us(void);
static void delay100us(void)
{
unsigned char time = 100;
while(--time);
}
void delay(unsigned char x,unsigned char y)
{
while(--x)
www.eeworm.com/read/239252/13292079
c delay.c
#include "DELAY.H"
static void delay100us(void);
static void delay100us(void)
{
unsigned char time = 100;
while(--time);
}
void delay(unsigned char x,unsigned char y)
{
while(--x)
www.eeworm.com/read/239206/13294516
cpp qcksrt.cpp
void qcksrt(int n, double arr[])
{
int m = 7; int nstack = 50; int fm = 7875; int fa = 211;
int fc = 1663; double a,fmi = 0.00012698413;
int istack[51];
int jstack = 0;
int i,j,
www.eeworm.com/read/239206/13294527
cpp sort2.cpp
void sort2(int n, double ra[], double rb[])
{
int l,ir,i,j;
double rra,rrb;
l = n / 2 + 1;
ir = n;
do
{
if (l > 1)
{
l = l - 1;
rra = ra[l];