我一直致力于循环调度计划。我的意见是：
Process Arrival Time Burst Time
1 0 4
2 2 2
3 4 3
4 6 5
5 7 1
时间片是3个单位！ 我的输出必须是：
Process AT BT WT TT FT
1 0 4 9 13 13
2 2 2 1 3 5
3 4 3 1 4 8
4 6 5 4 9 15
5 7 1 4 5 12
但是我没有得到过程1,4和＆amp;的正确结果（WT＆amp; FT）。 5.这是我的代码，任何人都可以帮我修复并获得上述结果吗？
#include<stdio.h>
#include<conio.h>
struct proc
{
int id;
int arrival;
int burst;
int rem;
int wait;
int finish;
int ti;
int turnaround;
float ratio;
}process[10];
int no,k;
int chkprocess(int);
void main()
{
int i,j,t,time = 0,n;
struct proc temp;
int nextprocess(int);
clrscr();
printf("\n \n Enter the number of processes: ");
scanf("%d", &n);
printf("\n \n Enter the time slice of the CPU: ");
scanf("%d", &t);
for(i = 1; i <= n; i++)
{
process[i].id = i;
printf("\n\nEnter the arrival time for process %d: ", i);
scanf("%d", &(process[i].arrival));
printf("\nEnter the burst time for process %d: ", i);
scanf("%d", &(process[i].burst));
process[i].rem = process[i].burst;
process[i].ti=0;
process[i].wait=0;
process[i].finish=0;
}
for(i = 1; i <= n; i++)
{
for(j = i + 1; j <= n; j++)
{
if(process[i].arrival > process[j].arrival)
{
temp = process[i];
process[i] = process[j];
process[j] = temp;
}
}
}
no = 0;
j = 1;
while(chkprocess(n) == 1)
{
if(process[no + 1].arrival == time)
no++;
if((process[j].ti<=t)&&(process[j].rem !=0))
{
process[j].rem--;
process[j].ti++;
for(i = 1; i <= no; i++)
{
if((i!=j) && (process[i].rem != 0))
process[i].wait++;
}
}
if(process[j].rem==0)
process[j].finish=time;
if((process[j].ti >= t)||(process[j].rem==0))
{
process[j].ti = 0;
j=nextprocess(j);
}
time++;
}
process[n].finish = time;
printf("\n\n Process Arrival Burst Waiting Finishing turnaround Tr/Tb \n");
printf("%5s %9s %7s %10s %8s %9s\n\n", "id", "time", "time", "time", "time", "time");
for(i = 1; i <= n; i++)
{
process[i].turnaround = process[i].wait + process[i].burst;
process[i].ratio = (float)process[i].turnaround / (float)process[i].burst;
printf("%5d %8d %7d %8d %10d %9d %10.1f ", process[i].id, process[i].arrival,
process[i].burst,
process[i].wait, process[i].finish,
process[i].turnaround, process[i].ratio);
printf("\n\n");
}
getch();
}
int chkprocess(int s)
{
int i;
for(i = 1; i <= s; i++)
{
if(process[i].rem != 0)
return 1;
}
return 0;
}
int nextprocess(int k)
{
int i;
i=k+1;
while(chkprocess(i) && i!=k)
{
if(process[i].rem != 0)
return i;
else
i=(i+1)%no;
}
}
谢谢
答案 0 :(得分：4)
我确信存在许多错误（如果用户想要输入11个或更多进程，即使您的进程数组限制为10个，也可以不关心。）
然而;我花了10分钟试图破译你的代码，但仍然不知道它认为它在做什么 - 根本没有评论，变量名和函数名没有帮助（例如no不是布尔值“是/否“变量，checkprocess()不检查一个进程，但检查所有进程以查看是否所有进程都已完成，等等）。大多数情况下，如果我得到报酬修复此代码，我会简单地将其丢弃并从头开始重写以节省时间。我想过从头开始重写它，只是发布结果代码;但这不会帮助你完成你的作业。
我的建议是，从头开始重写，而不是修复它。
它应该有一个全局currently_running_process变量，一个全局current_time变量，一个用于增加当前时间的函数，以及一个用于调度程序本身的函数。
增加当前时间的功能是：
current_time++ current_time == arrival_time）并将任何已启动的进程附加到调度程序链接列表的末尾调度程序功能应该：
increase_time()函数，直到经过了这段时间注意：我从current_time = -1;开始并在调用调度程序函数之前调用该函数以增加当前时间;这样在调度程序开始工作之前，任何带有arrival_time == 0的进程都会被添加到调度程序的链表中（并且调度程序函数一旦启动就不会看到空列表）。
答案 1 :(得分：2)
/* The following code doesn't take the arrival time of the processes in account.
HAPPY CODING */
#include<stdio.h>
void main()
{
int b[10],br[10],wo[10];
int n,i,bt,q,count;
float awt=0,att=0;
for (i=0;i<10;i++)
wo[i]=0;
printf("Input the nmbr of processes running....");
scanf("%d",&n);
printf("\n Input their burst tym in order..");
for(i=0;i<n;i++)
scanf("%d",&b[i]);
printf("\n Input the quantum time for the algorithm..");
scanf("%d",&q);
for(i=0;i<n;i++)
br[i]=b[i];
bt=0;
for(i=0;i<n;i++)
bt=bt+b[i];
count=0;
printf("\nThe Gantt Chart is as follows:\n");
printf("\n 0");
do
{
for(i=0;i<n;i++)
{
if(br[i]==0)
{}
else
{
if(br[i]>=q)
{
br[i]=br[i]-q;
if(br[i]==0)
wo[i]=count;
count=count+q;
printf("\t(P%d)",i);
printf("\t%d",count);
}
else
{
if(br[i]<q)
{
count=count+br[i];
br[i]=0;
wo[i]=count;
printf("\t(P%d)",i);
printf("\t%d",count);
}
}
}
}
}while(count<bt);
for(i=0;i<n;i++)
awt=awt+(wo[i]-b[i]);
awt=awt/n;
printf("\n The average waiting time is....%f",awt);
for(i=0;i<n;i++)
att=att+wo[i];
att=att/n;
printf("\n The average turnaround time is....%f",att);
}
答案 2 :(得分：1)
你可以使用队列做同样的事情，我粘贴一个用ANSI CPP编写的链接 您可以查看此链接以获取更多信息。我遇到了和你一样的问题，但链接上的代码对我帮助很大，它还包含许多其他调度程序，但我只提取了循环法。 click here to see the code for round robin Scheduling
答案 3 :(得分：0)
#include<stdio.h>
#include<conio.h>
main(){
int i, j, k, n, so, tq, sob, sum, swt, stat, tata, temp, count;
int bt[10], bth[10], wt[10], tat[10];
float awt=0.0, atat=0.0;
char new;
// i = loop controller
// j = loop controller
// k = loop controller
// n = number of process
// so = (burst time holder divided by time quantum) and added by one
// tq = time quantum
// awt =average waiting time
// new = hold the value of start command
// sob = gantt chart size from so
// swt = summation of waiting time l
// bt[] = burst time
// wt[] = waiting time
// atat = average turn around time
// gcps = gantt chart process sequence
// stat = summation of turn around time
// tata = accumulator of turn around time
// temp = time quantum holder
// count = counter
// bth[] = burst time holder
// tat[] = turn around time
printf("\n\n\n\n To start round robin scheduling press any key: ");
k = 0;
new = getche();
system("cls");
while(k < 7){
j = 0; sob = 0; count = 0; sum = 0; swt = 0; stat = 0; tata = 0;
printf("\n\n\n\t\t\t ROUND-ROBIN SCHEDULING");
printf("\n\t\t\t ======================");
printf("\n\n\n\n\n Enter number of processes: ");
scanf("%d", &n);
printf("\n");
for(i = 0; i < n; i++){
printf("\n Enter burst time for Process P%d: ", i+1);
scanf("%d", &bt[i]);
bth[i] = bt[i];
}
printf("\n\n Enter time quantum: ");
scanf("%d", &tq);
system("cls");
printf("\n\n\n\t\t\t ROUND-ROBIN SCHEDULING");
printf("\n\t\t\t ======================");
printf("\n\n\n\n\n Time quantum: %d", tq);
for(i = 0; i < n; i++){
if(bth[i] % tq == 0){
so = bth[i] / tq;
}
else{so = (bth[i] / tq) +1;}
sob = sob + so;
}
int gc[sob], gcps[sob];
while(1){
for(i = 0,count = 0; i < n; i++){
temp = tq;
if(bth[i] == 0){
count++;
continue;
}
if(bth[i] > tq){
gc[j] = tq;
gcps[j] = i+1; j++;
bth[i] = bth[i] - tq;
}
else if(bth[i] >= 0){
if(bth[i] == tq){gc[j] = tq; gcps[j] = i+1; j++;}
else{gc[j] = bth[i]; gcps[j] = i+1; j++;}
temp = bth[i];
bth[i] = 0;
}
tata = tata + temp;
tat[i ]= tata;
}
if(n==count){
break;
}
}
for(i = 0; i < n; i++){
wt[i] = tat[i] - bt[i];
swt = swt + wt[i];
stat = stat + tat[i];
}
awt = (float)swt/n;
atat = (float)stat/n;
printf("\n\n Process Burst time Waiting time Turn around time\n");
printf(" ------- ---------- ------------ ----------------\n");
for(i = 0; i < n; i++){
printf("\n\n P%d\t %d\t %d \t %d", i+1, bt[i], wt[i], tat[i]);
}
printf("\n\n\n\n Gantt Chart:\n");
printf(" ------------\n\n");
for(j = 0; j < sob; j++){
printf("\tP%d", gcps[j]);
}
printf("\n 0");
for(j = 0; j < sob; j++){
sum = sum + gc[j];
if(j == 0){printf(" %d", sum);}
else{printf("\t %d", sum);}
}
printf("\n\n\n\n Average waiting time: %.2f \n\n Average turn around time: %.2f",awt,atat);
printf("\n\n\n\n To start again press S and to exit press any key: ");
new = getche();
system("cls");
if(new == 'S'|| new == 's'){k++;}
else{printf("\n\n\n Program was terminated successfully\n\n Thank you\n\n\n"); break;}
}
}