C語言實現OPT、FIFO及LRU等頁面置換算法

假設有10個頁面，n個頁框。頁面的訪問順序為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。

當n在[1,10]中取值時，請編寫程序實現OPT、LRU、FIFO頁面置換算法，並根據頁面訪問順序模擬執行，分別計算缺頁數量。

1.1思路:

• FIFO:采用隊列存儲,隊列最大容量可變,設為n.

訪問->未找到(缺頁數++)->嘗試將缺頁加入隊列->容量夠則加入隊尾,否則出隊首元素,並將新元素加入隊尾（即順序前移）.

• LRU:鏈表法實現,鏈表最大長度為n

訪問:1.未找到(缺頁數++)->嘗試將缺頁加入鏈表->容量夠則加入鏈表頭,否則淘汰鏈表尾,並加入鏈表頭部

2.找到->將對應鏈表節點提到頭節點.

• OPT:未來最久不被使用的頁面

訪問->未找到(缺頁數++)->嘗試將缺頁加入頁框->容量夠則加入(數組),否則,計算當前時刻頁框中所有頁面距離下一次使用的時間,取最大的淘汰,並加入新的.

1.2代碼實現：

#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];//隊列
    int head = 0;//隊首 
    int tail = 0;//隊尾 
    int sum = 0;//缺頁數 
    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) {
            //沒找到
            sum++; 
            if (tail < n) {
                //未滿，加入隊尾 
                table[tail++] = tar;
            } else {
                for (k = head; k+1 < tail; k++) {
                    table[k] = table[k+1];
                    //順序前移 
                }
                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) {
                //就在頭節點，無需操作 
            } 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) {
                //容量滿了
                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;
}