我正在研究Shortest Remaining Time Next Scheduling，我必须检查每1个时间单位，看看是否有另一个剩余时间较短的作业，如果相等则保持当前进程。对于我使用的输入：
PID ArrivalTime Burst/ExecutionTime
1 0 6
2 3 2
3 5 1
4 9 7
5 10 5
6 12 3
7 14 4
8 16 5
9 17 7
10 19 2
和我的输出:(左边是我得到的，右边是它应该是什么）：
PID WAIT TURNAROUND PID WAIT TURNAROUND
1 0 9 1 0 9
2 0 2 2 0 2
3 0 1 3 0 1
4 0 26 4 0 26
5 0 14 5 0 5
6 0 3 6 3 6
7 1 5 7 4 10
8 8 13 8 8 13
9 18 25 9 18 25
10 0 2 10 0 2
Average Wait: 2.7 Ave Turnaround 10.0 Average Wait: 3.3 Average Turnaround 9.9
我无法缩小srtn函数中问题所在的范围，除了我的3个输出都是正确的，这更令人困惑！任何帮助将不胜感激！
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <termios.h>
#include <signal.h>
#include <fcntl.h>
#include <sys/types.h>
#define LINELEN 512
#define MAX_PROCESS 100
#define TIME_QUANTUM 1
typedef struct process
{
int ID;
int arrival_time;
int time_to_completion;
double wait_time;
double turn_around;
double time_wait;
int active;
}process;
void fcfs(struct process[MAX_PROCESS], int);
void sjf (struct process[MAX_PROCESS], int);
void srtn(struct process[MAX_PROCESS], int);
void rr (struct process[MAX_PROCESS], int);
void rrc(struct process[MAX_PROCESS], int);
void print_info(struct process[MAX_PROCESS], int);
void sort_by_time(struct process array[MAX_PROCESS], int num_valid_pid);
int main(int ac,char *av[])
{
int counter=0;
int p1=0, p2=0, p3=0;
process array[MAX_PROCESS];
while ( scanf("%d %d %d", &p1, &p2, &p3) != EOF ){//Get all the info available and put it in array of structs
array[counter].ID = p1;
array[counter].arrival_time = p2;
array[counter].time_to_completion = p3;
array[counter].active = 0;
counter++;
}
//fcfs(array, counter);
//sjf (array, counter);
srtn(array, counter);
/*rr (array, counter);
rrc(array, counter);*/
return 0;
}
void srtn(struct process array[MAX_PROCESS], int num_pid){
printf("Shortest Remaining Time Next\n");//for the output so we know what algorithm
//create an array of pids that are valid to search.
int num_valid_processes = 0, current_time=0, i,j, next_process, counter = 0, fin_pid = 0, keep_going=0;//declarations
process to_sort[MAX_PROCESS];
//we want to do this next loop for as many processes as we have, or num_pid
while(keep_going!=1){
//adds all the available processes to the to sort array to be sorted
//available means that it has arrived, which means it is <= current_time
//after it gets all the processes, it breaks out of the for loop
for(i=counter; i<num_pid; i++){
if(array[i].arrival_time<=current_time){
to_sort[num_valid_processes]=array[i];
num_valid_processes++;
counter++;
}
else
break;
}
//sort the to_sort array by the time_to_completion
sort_by_time(to_sort,num_valid_processes);
//set the wait time and turnaround time for the next process
next_process = to_sort[0].ID-1;
if(array[next_process].active==0){//the id hasn't had the wait time calculated yet
array[next_process].wait_time = current_time-array[next_process].arrival_time;
array[next_process].active=1;
array[next_process].time_wait = current_time;
}
if(array[next_process].time_to_completion <= TIME_QUANTUM){
array[next_process].turn_around = array[next_process].wait_time + (current_time-array[next_process].time_wait)+array[next_process].time_to_completion;
fin_pid++;
//delete the process we just worked on so we don't get duplicates.
num_valid_processes--;
for(i=0;i<num_valid_processes;i++){
to_sort[i]=to_sort[i+1];
}
}
else{
array[next_process].time_to_completion = array[next_process].time_to_completion - TIME_QUANTUM;
//to_sort[0].time_to_completion = to_sort[next_process].time_to_completion - TIME_QUANTUM;
}
current_time = current_time+TIME_QUANTUM;
if(fin_pid==num_pid)
keep_going=1;
}
print_info(array, num_pid);
}
void print_info(struct process array[MAX_PROCESS], int num_pid){
int i;
double tot_wait=0, tot_turn = 0;
printf("\x1b[04mPID\tWAIT\tTURNAROUND\n\x1b[24m");
for(i=0; i<num_pid; i++){
printf("%d\t%.0f\t%.0f\n", array[i].ID, array[i].wait_time, array[i].turn_around);
tot_wait=tot_wait+array[i].wait_time;
tot_turn = tot_turn +array[i].turn_around;
}
printf("Average Wait: %.1f Average Turnaround %.1f\n", tot_wait/num_pid, tot_turn/num_pid);
}
void sort_by_time(struct process array[MAX_PROCESS], int num_valid_pid)
{
int i,j;
for (i = 0; i < num_valid_pid; i++)
{
int min = i;
for (j = i+1; j < num_valid_pid; j++){
if (array[j].time_to_completion < array[min].time_to_completion)
min = j;
if (array[j].time_to_completion == array[min].time_to_completion){
if(array[j].ID<array[min].ID)
min = j;
}
}
process temp = array[i];
array[i] = array[min];
array[min] = temp;
}
}
答案 0 :(得分：0)
问题发生在时间12 - pid 6显示需要3秒而pid 5正在运行3秒。如何解决需要相同剩余时间的进程之间的联系？赞成pid 6会给你左边的结果，而pid 5会给你右边的结果。鉴于你的问题定义很弱，要么是正确的。