操作系統(tǒng)_頁(yè)面置換算法FIFO,OPT,LRU實(shí)現(xiàn)

1、操作系統(tǒng)第六次上機(jī)在一個(gè)請(qǐng)求分頁(yè)系統(tǒng)中，設(shè)頁(yè)面大小占 100個(gè)單元，假如系統(tǒng)分配給一個(gè)作業(yè)的物理塊數(shù)為3，試求出用FIFO , LRU，OPT三種算法在程序訪問(wèn)過(guò)程中所發(fā)生的缺頁(yè)次數(shù)及缺頁(yè)率，每次中斷時(shí)都需要打印出來(lái)或者標(biāo)示出來(lái)。（假設(shè)最初頁(yè)面都在外存）1. 假定此作業(yè)的訪問(wèn)地址序列為202，313，252，111，546，217，444，544，365，223，398，111。2. 輸入任意的訪問(wèn)序列，也必須正確顯示。/*代碼尚需完善：1應(yīng)由用戶輸入數(shù)組，且應(yīng)根據(jù)題目要求對(duì)每個(gè)數(shù) /100，得到頁(yè)塊標(biāo)號(hào) 2在動(dòng)態(tài)輸入的情況下，通過(guò)sizeof,獲得數(shù)組長(zhǎng)度，實(shí)現(xiàn)任意輸入的處理3. FIFO算

2、法實(shí)現(xiàn)，why ？4. 在OPT實(shí)現(xiàn)中，mark屬性設(shè)置，以及向后遍歷的參數(shù)設(shè)置？*/前三個(gè)頁(yè)塊單獨(dú)處理，需注意前三個(gè)頁(yè)塊也可能重復(fù)。已做出修改!#in clude<iostream> using n amespace std; intin put12 = 2,3,2,1,5,2,4,5,3,2,3,1 ;class pagepublic:int num;int mark;page()num = 0;mark = -1;void FIFO()cout << "FIFO"<< en dl;int error = 0;page frame3;

3、/ 頁(yè)幀bool flag = true;int check = 0;for (int i = 0; i<3; i+)/處理前三個(gè)引用for (int k = 0; k < i; k+) if (in puti = in putk)flag = false;if (flag = true) framei. num = in puti;framei.mark = i;erro r+;cout << framei. num << " | "for (i nt j = 0; j <= i; j+)cout << framej.

4、 num << ':cout << en dl;elsecheck+;for (int i = 3-check; i<12; i+)int j;for (j = 0; j<3; j+)if (in puti = frame。, num)cout << in puti << en dl;break;if (j = 3)erro r+;/換掉最舊的頁(yè)frame(error - 1) % 3) .num = in puti;/ ?cout << in puti << " | "for (i

5、nt k = 0; k<3; k+)cout << framek, num << ':cout << en dl;cout << "FIFO:" << endl;cout << "Error 次數(shù):"<< error << en dl;cout << "Frame Error:" << (error/12.0) << endl << en dl;void OPT()cout &l

6、t;< "OPT"<< en dl;int error = 0;page frame3;bool flag = true;int check = 0;for (int i = 0; i<3; i+)/處理前三個(gè)引用for (int k = 0; k < i; k+) if (in puti = in putk)flag = false;if (flag = true) framei. num = in puti;erro r+;cout << framei. num << " | "for (i nt

7、 j = 0; j <= i; j+)cout << framej. num << ''cout << en dl;elsecheck+;for (int i = 3-check; i<12; i+)int j;for (j = 0; j<3; j+)if (in puti = frame。, num)cout << in puti << en dl; break;if (j = 3)erro r+;for (j = 0; j<3; j+)framej.mark = 21;/ ?for (int

8、 k =20; k >= i; k-)/向后遍歷，找到最長(zhǎng)時(shí)間不用的頁(yè)k =20 ?貌似修改后會(huì)岀問(wèn)題if (frame。, num = in putk) framej.mark = k;if (frame0.mark>frame1.mark&&frame0.mark>frame2.mark) frame0, num = in puti;else if (frame1.mark>frame0.mark&&frame1.mark>frame2.mark) frame1. num = in puti;elseframe2. num =

9、in puti;cout << in puti << " | "for (int k = 0; k<3; k+)cout << framek. num << ':cout << en dl;cout << "OPT:" << endl;cout << "Error 次數(shù):"<< error << en dl;cout << "Frame Error:" <<

10、(error / 12.0) << endl << en dl;void LRU()cout << "LRU"<< en dl;int error = 0;page frame3;bool flag = true;int check = 0;for (i nt i = 0; i<3; i+)處理前三個(gè)引用for (int k = 0; k < i; k+) if (in puti = in putk)flag = false;if (flag = true) framei. num = in puti;erro r

11、+;cout << framei. num << " | "for (i nt j = 0; j <= i; j+)cout << framej. num << ''cout << en dl;elsecheck+;for (int i = 3-check; i<12; i+)int j;for (j = 0; j<3; j+)if (in puti = frame。, num)cout << in puti << en dl;break;if (j = 3

12、)erro r+;for (j = 0; j<3; j+)framej.mark = -1;for (i nt k = 0; k <= i; k+)向前遍歷，找到最近最少使用的if (framej, num = in putk)framej.mark = k;if (frame0.mark<frame1.mark&&frame0.mark<frame2.mark)frame0. num = in puti;else if (frame1.mark<frame0.mark&&frame1.mark<frame2.mark)frame1. num = in puti;elseframe2. num = in puti;cout << in puti << " | "for (int k = 0; k<3; k+)cout << framek. num << ':cout << en dl;cout << "LRU:" << endl;cou