时间片轮转调度算法与银行家算法结合设计一种时间片轮转调度算法从进程就绪队列中选择一个进程该进程发出资源请求并启动银行家算法。用C语言实现。
以下是一个基于时间片轮转调度算法与银行家算法结合的简单示例代码:
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#define MAX_PROCESSES 10
#define MAX_RESOURCES 3
#define TIME_QUANTUM 2
int available[MAX_RESOURCES];
int maximum[MAX_PROCESSES][MAX_RESOURCES];
int allocation[MAX_PROCESSES][MAX_RESOURCES];
int need[MAX_PROCESSES][MAX_RESOURCES];
bool finished[MAX_PROCESSES] = {false};
struct process {
int id;
int arrival_time;
int burst_time;
int remaining_time;
int waiting_time;
int turnaround_time;
int resources[MAX_RESOURCES];
};
int num_processes = 0;
struct process processes[MAX_PROCESSES];
int get_safe_process() {
int work[MAX_RESOURCES];
bool finish[num_processes];
int safe_sequence[num_processes];
int num_finished = 0;
int i, j;
for (i = 0; i < MAX_RESOURCES; i++) {
work[i] = available[i];
}
for (i = 0; i < num_processes; i++) {
finish[i] = finished[i];
}
while (num_finished < num_processes) {
bool found = false;
for (i = 0; i < num_processes; i++) {
if (!finish[i]) {
bool can_finish = true;
for (j = 0; j < MAX_RESOURCES; j++) {
if (need[i][j] > work[j]) {
can_finish = false;
break;
}
}
if (can_finish) {
found = true;
safe_sequence[num_finished] = i;
num_finished++;
finish[i] = true;
for (j = 0; j < MAX_RESOURCES; j++) {
work[j] += allocation[i][j];
}
}
}
}
if (!found) {
return -1;
}
}
return safe_sequence[0];
}
void execute_process(int i, int time) {
int j;
for (j = 0; j < MAX_RESOURCES; j++) {
available[j] -= processes[i].resources[j];
allocation[i][j] += processes[i].resources[j];
need[i][j] = maximum[i][j] - allocation[i][j];
}
processes[i].remaining_time = processes[i].burst_time - time;
if (processes[i].remaining_time == 0) {
processes[i].turnaround_time = time - processes[i].arrival_time;
processes[i].waiting_time = processes[i].turnaround_time - processes[i].burst_time;
finished[i] = true;
for (j = 0; j < MAX_RESOURCES; j++) {
available[j] += allocation[i][j];
allocation[i][j] = 0;
need[i][j] = 0;
}
}
}
void print_processes() {
int i, j;
printf(" ID Arrival Burst Remaining Waiting Turnaround\n");
for (i = 0; i < num_processes; i++) {
printf("%5d%8d%7d%11d%9d%11d\n", processes[i].id, processes[i].arrival_time, processes[i].burst_time, processes[i].remaining_time, processes[i].waiting_time, processes[i].turnaround_time);
printf(" Maximum: ");
for (j = 0; j < MAX_RESOURCES; j++) {
printf("%2d ", maximum[i][j]);
}
printf("\n Allocation: ");
for (j = 0; j < MAX_RESOURCES; j++) {
printf("%2d ", allocation[i][j]);
}
printf("\n Need: ");
for (j = 0; j < MAX_RESOURCES; j++) {
printf("%2d ", need[i][j]);
}
printf("\n");
}
}
int main() {
int i, j;
printf("Enter the number of processes: ");
scanf("%d", &num_processes);
printf("Enter the available resources: ");
for (i = 0; i < MAX_RESOURCES; i++) {
scanf("%d", &available[i]);
}
for (i = 0; i < num_processes; i++) {
printf("Enter the arrival time and burst time for process %d: ", i);
scanf("%d%d", &processes[i].arrival_time, &processes[i].burst_time);
processes[i].id = i;
processes[i].remaining_time = processes[i].burst_time;
for (j = 0; j < MAX_RESOURCES; j++) {
printf("Enter the maximum number of resources %d requires of type %d: ", i, j);
scanf("%d", &maximum[i][j]);
processes[i].resources[j] = 0;
allocation[i][j] = 0;
need[i][j] = maximum[i][j];
}
}
printf("\nTime Process\n");
int time = 0;
int current_process = -1;
while (true) {
int i;
for (i = 0; i < num_processes; i++) {
if (!finished[i] && processes[i].arrival_time <= time) {
current_process = i;
break;
}
}
if (current_process == -1) {
printf("%4d Idle\n", time);
time++;
continue;
}
int remaining_time = processes[current_process].remaining_time;
if (remaining_time > TIME_QUANTUM) {
processes[current_process].remaining_time -= TIME_QUANTUM;
printf("%4d %d\n", time, current_process);
execute_process(current_process, TIME_QUANTUM);
} else {
printf("%4d %d\n", time, current_process);
execute_process(current_process, remaining_time);
current_process = -1;
}
int safe_process = get_safe_process();
if (safe_process != -1) {
current_process = safe_process;
}
if (current_process == -1) {
int num_unfinished = 0;
for (i = 0; i < num_processes; i++) {
if (!finished[i]) {
num_unfinished++;
}
}
if (num_unfinished == 0) {
break;
}
}
time++;
}
print_processes();
return 0;
}
该程序首先要求用户输入进程数和可用资源数,然后输入每个进程的到达时间、执行时间和各种资源的最大需求量。然后程序模拟时间片轮转调度算法的执行过程,每个时间片执行一个进程(如果有),直到所有进程完成。在每个时间片中,程序执行以下步骤:
- 查找就绪队列中最早到达的进程。
- 如果没有就绪的进程,则输出“Idle”。
- 如果当前执行的进程剩余执行时间大于时间片长度,则执行进程并减少其剩余执行时间。
- 如果当前执行的进程剩余执行时间小于等于时间片长度,则执行进程并将其标记为已完成。
- 如果当前执行的进程已完成,则释放其占用的所有资源。
- 使用银行家算法检查系统是否处于安全状态。
- 如果系统处于安全状态,则选择下一个要执行的进程。
- 如果系统不处于安全状态,则继续执行当前进程。
- 如果所有进程都已完成,则退出循环。
- 输出所有进程的等待时间和周转时间。
该程序使用了一个结构体来表示每个进程,并使用了多维数组来存储资源的需求、分配和可用情况。在每个时间片中,程序使用了银行家算法来检查系统是否处于安全状态,并根据检查结果决定下一个要执行的进程。最后,程序输出了每个进程的等待时间和周转时间,以及其最大资源需求、分配和需求的详细信息
原文地址: https://www.cveoy.top/t/topic/g8yO 著作权归作者所有。请勿转载和采集!