C语言实现OPT、FIFO及LRU等页面置换算法

假设有10个页面，n个页框。页面的访问顺序为0, 9, 8, 4, 4, 3, 6, 5, 1, 5, 0, 2, 1, 1, 1, 1, 8, 8, 5,

3, 9, 8, 9, 9, 6, 1, 8, 4, 6, 4, 3, 7, 1, 3, 2, 9, 8, 6, 2, 9, 2, 7, 2, 7, 8, 4, 2, 3, 0, 1, 9, 4,

7, 1, 5, 9, 1, 7, 3, 4, 3, 7, 1, 0, 3, 5, 9, 9, 4, 9, 6, 1, 7, 5, 9, 4, 9, 7, 3, 6, 7, 7, 4, 5, 3, 5, 3, 1, 5, 6, 1, 1, 9, 6, 6, 4, 0, 9, 4, 3。

当n在[1,10]中取值时，请编写程序实现OPT、LRU、FIFO页面置换算法，并根据页面访问顺序模拟执行，分别计算缺页数量。

1.1思路:

• FIFO:采用队列存储,队列最大容量可变,设为n.

访问->未找到(缺页数++)->尝试将缺页加入队列->容量够则加入队尾,否则出队首元素,并将新元素加入队尾（即顺序前移）.

• LRU:链表法实现,链表最大长度为n

访问:1.未找到(缺页数++)->尝试将缺页加入链表->容量够则加入链表头,否则淘汰链表尾,并加入链表头部

2.找到->将对应链表节点提到头节点.

• OPT:未来最久不被使用的页面

访问->未找到(缺页数++)->尝试将缺页加入页框->容量够则加入(数组),否则,计算当前时刻页框中所有页面距离下一次使用的时间,取最大的淘汰,并加入新的.

1.2代码实现：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define tot 100
int schedule[tot] = {0, 9, 8, 4, 4, 3, 6, 5, 1, 5, 0, 2, 1, 1, 1, 1, 8, 8, 5, 

3, 9, 8, 9, 9, 6, 1, 8, 4, 6, 4, 3, 7, 1, 3, 2, 9, 8, 6, 2, 9, 2, 7, 2, 7, 8, 4, 2, 3, 0, 1, 9, 4,

7, 1, 5, 9, 1, 7, 3, 4, 3, 7, 1, 0, 3, 5, 9, 9, 4, 9, 6, 1, 7, 5, 9, 4, 9, 7, 3, 6, 7, 7, 4, 5, 3, 5, 3, 1, 5, 6, 1, 

1, 9, 6, 6, 4, 0, 9, 4, 3};
int FIFO(int n) {
    int table[n];//队列
    int head = 0;//队首 
    int tail = 0;//队尾 
    int sum = 0;//缺页数 
    int i, j, k;
    int flag = 0;
    
    for (i = 0; i < tot; i++) {
        int tar = schedule[i];
        flag = 0;
        for (j = head; j < tail; j++) {
            if (tar == table[j]) {
                //找到
                flag = 1;
                break;
            }
        }
        if (!flag) {
            //没找到
            sum++; 
            if (tail < n) {
                //未满，加入队尾 
                table[tail++] = tar;
            } else {
                for (k = head; k+1 < tail; k++) {
                    table[k] = table[k+1];
                    //顺序前移 
                }
                table[tail-1] = tar;
            }
        }
    }
    return sum;
}

typedef struct Node{
    int order;
    struct Node * next;
}Node, *NPT;
int LRU(int n) {
    Node node[n];
    NPT head = NULL, p, r;
    int i, j, k, cnt = 0;
    int sum = 0;
    int flag = 0;
    
    for (i = 0; i < tot; i++) {
        flag  = 0; 
        int tar = schedule[i];
        r = NULL;
        p = head;
        while (p != NULL) {
            if (p->order == tar) {
                //找到
                flag = 1;
                break; 
            } 
            r = p;
            p = p->next;
        } 
        if (flag) {
            //找到 
            if (r == NULL) {
                //就在头节点，无需操作 
            } else {
                r->next = p->next;
                p->next = head;
                head = p;
            } 
        } else {
            sum++;
            p = (NPT) malloc(sizeof(Node));
            p->order = tar;
            p->next = NULL;
            if (cnt == n) {
                //容量限制
                r = head;
                if (r->next == NULL) {
                    head = p;
                } else {
                    while (r->next->next != NULL) {
                        r = r->next;
                    }
                    r->next = NULL;
                    p->next = head;
                    head = p;
                } 
            } else {
                p->next = head;
                head = p;
                cnt++;
            }
        } 
    }
    return sum;
}

int OPT(int n) {
    int table[n];
    int cnt = 0, sum = 0;
    int i, j, k, flag = 0;
    for (i = 0; i < tot; i++) {
        int tar = schedule[i];
        flag = 0;
        for (j = 0; j < cnt; j++) {
            if (table[j] == tar) {
                flag = 1;
                break;
            }
        }
        if (flag) {
            //找到
             
        } else {
            sum++;
            if (cnt == n) {
                //容量满了
                int max = 0;
                int index = -1;
                int temp = 0;
                for (j = 0; j < cnt; j++) {
                    temp = 0;
                    for (k = i + 1; k < tot; k++) {
                        if (table[j] == schedule[k]) {
                            temp++;
                            break;
                        } else temp++;
                    }
                    if (k == tot) {
                        temp = 1000;
                    }
                    if (temp > max) {
                        max = temp;
                        index = j;
                    }
                } 
                table[index] = tar;
            } else {
                table[cnt++] = tar;
            }
        }
    }
    return sum;
}

int main(){
    int i;
    for (i = 1; i <= 10; i++) {
        printf("OPT= %d, FIFO = %d, lru = %d\n",OPT(i), FIFO(i), LRU(i));
    }
    return 0;
}