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/315422/13543649
prg 胆码测试程序.prg
SET TALK OFF
DIME S(30)
DIME BB(5)
DIME BS(5)
DIME BG(5)
T="●"
F="×"
N=1
h1=0
h2=0
h3=0
SELE A
USE X3D.DBF
SELE J
USE 胆码测试表.dbf
dele all
pack
sele a
do while .NOT.EOF()
S(1)=
www.eeworm.com/read/315285/13547196
c b.c
#include
struct f
{
char filename[8];
char fileex[3];
char shuxing;
char baoliu[10];
char shijian[2];
char riqi[2];
char qishi[2];
char daxiao[4];
}file[244];
int fd_rename(char *oldf
www.eeworm.com/read/315110/13551501
c safeio.c
/* Copyright (C) 1999 Drazen Kacar
This file is part of LibGTop 1.0.
Contributed by Drazen Kacar , May 1999.
LibGTop is free software; you can redistribute it and/or modify it
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txt lianjie.txt
【程序80】
题目：海滩上有一堆桃子，五只猴子来分。第一只猴子把这堆桃子凭据分为五份，多了一个，这只
猴子把多的一个扔入海中，拿走了一份。第二只猴子把剩下的桃子又平均分成五份，又多了
一个，它同样把多的一个扔入海中，拿走了一份，第三、第四、第五只猴子都是这样做的，
问海滩上原来最少有多少个桃子？
1.程序分析：
2.程序源代码：
main()
{int i,m,j ...
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txt studentsystem.txt
#include "stdio.h"
#include "stdlib.h"
#include "string.h"
int shoudsave=0;
struct student
{
char num[10];//学号
char name[20];
char sex[4];
int cgrade;
www.eeworm.com/read/314531/13564903
cpp wfyb.cpp
#include
#include
#include
void youbu(int Ruleright[][7],int Link[][8])
{
bool First[34][36]={0};
bool Follow[34][36]={0};
int Rtrace,Rptr;
int Ltrace,Lp
www.eeworm.com/read/314531/13564905
cpp table.cpp
#include"bianyi.h"
#define N 21
#define M 50
#define SIZE 36
struct nmdzb_type //内码对照表的数组
{
int stringsnb;
char strings[20];
}nmb[SIZE]={{0," "},{1," "},{2," "},{3,"PROG
www.eeworm.com/read/314202/13571635
c mcu_uart.c
/* Copyright (C) 1996-2005 Brilliant Ideal Electronics. All rights reserved.
MP3_Player+USB_Disk V3.0 Edit by JMBIE STUDIO 2005.03
*/
#include "AT89C51SND1_REG.H"
#include "MCU_UART.H"
ch
www.eeworm.com/read/314126/13574204
c text1.c
#include
unsigned char i;
unsigned char temp;
unsigned char a,b;
void delay(void)
{
unsigned char m,n,s;
for(m=20;m>0;m--)
for(n=20;n>0;n--)
www.eeworm.com/read/314126/13574274
c text1.c
#include
#include
sbit led0=P0^0;
sbit led1=P0^1;
sbit led2=P0^2;
sbit led3=P0^3;
sbit led4=P0^4;
sbit led5=P0^5;
sbit led6=P0^6;
sbit led7=P0^7;
int i,j;
void main()
{