写一个C语言代码采用最近最久未使用置换算法LRU。输入给定的页面访问序列和内存的物理块数下,输出缺页率。 二、实验测试: 【输入】 虚页的个数N:8 //(虚页序号0--7) 虚页访问次数num:20 页面访问序列:7 0

以下是采用最近最久未使用置换算法LRU的C语言代码：

#include <stdio.h>
#include <stdlib.h>

#define MAX 100

int main()
{
    int N, num, i, j, k, page_faults = 0;
    int pages[MAX], frames[MAX], counter[MAX];
    int flag1, flag2, pos, max;

    printf("Enter the number of pages: ");
    scanf("%d", &N);

    printf("Enter the number of page accesses: ");
    scanf("%d", &num);

    printf("Enter the page reference string: ");
    for(i=0; i<num; i++)
        scanf("%d", &pages[i]);

    printf("Enter the number of physical memory frames: ");
    scanf("%d", &k);

    for(i=0; i<k; i++)
    {
        frames[i] = -1;
        counter[i] = 0;
    }

    for(i=0; i<num; i++)
    {
        flag1 = flag2 = 0;

        for(j=0; j<k; j++)
        {
            if(frames[j] == pages[i])
            {
                counter[j] = 1;
                flag1 = flag2 = 1;
                break;
            }
        }

        if(flag1 == 0)
        {
            for(j=0; j<k; j++)
            {
                if(frames[j] == -1)
                {
                    page_faults++;
                    frames[j] = pages[i];
                    counter[j] = 1;
                    flag2 = 1;
                    break;
                }
            }
        }

        if(flag2 == 0)
        {
            for(j=0; j<k; j++)
                counter[j]++;

            max = counter[0];
            pos = 0;

            for(j=1; j<k; j++)
            {
                if(counter[j] > max)
                {
                    max = counter[j];
                    pos = j;
                }
            }

            frames[pos] = pages[i];
            counter[pos] = 1;
            page_faults++;
        }

        printf("Page reference string: ");
        for(j=0; j<=i; j++)
            printf("%d ", pages[j]);

        printf("\nFrame Status: ");
        for(j=0; j<k; j++)
        {
            if(frames[j] == -1)
                printf("X ");
            else
                printf("%d ", frames[j]);
        }

        printf("\n");
    }

    printf("Total number of page faults: %d\n", page_faults);
    printf("Page Fault Rate: %.2f\n", (float)page_faults/num);

    return 0;
}

输入示例：

Enter the number of pages: 8
Enter the number of page accesses: 2
Enter the page reference string: 7 0
Enter the number of physical memory frames: 3

输出示例：

Page reference string: 7
Frame Status: 7 X X
Page reference string: 7 0
Frame Status: 7 0 X
Total number of page faults: 2
Page Fault Rate: 1.00