# [2020q3](http://wiki.csie.ncku.edu.tw/sysprog/schedule) 第 12 週測驗題 ###### tags: `sysprog2020` :::info 目的: 檢驗學員對 [CS:APP 第 9 章](https://hackmd.io/@sysprog/CSAPP-ch9)和[高效 Web 伺服器開發](https://hackmd.io/@sysprog/fast-web-server)的認知 ::: ==[作答表單](https://docs.google.com/forms/d/e/1FAIpQLSet46ZaQp-ZnqnuvjX9_BNdYo1x1jTzfYzElLegews6LXIXCQ/viewform)== --- ### 測驗 `1` 以下關於記憶體管理的描述,選出其中唯一正確的描述 ==作答區== Q1 = ? * `(a)` 在一個 [buddy system](https://en.wikipedia.org/wiki/Buddy_system) 中,最高可達 50% 的空間可因內部碎片化 (fragmentation) 而被浪費 * `(b)` 平均來說,first-fit 演算法會比 best-fit 演算法來得慢 * `(c)` 僅在 free list 依據記憶體地址遞增排序時,使用邊界標記來回收 (reclaim) 才會快速 * `(d)` [buddy system](https://en.wikipedia.org/wiki/Buddy_system) 只會有內部碎片,不會有外部碎片 --- ### 測驗 `2` 以下關於記憶體管理的描述,選出其中唯一正確的描述 ==作答區== Q2 = ? * `(a)` 在按照 block size 遞減排序的 free list 中,使用 first-fit 演算法會導致配置的效能低落,但可避免外部碎片 * `(b)` 對於 best-fit 演算法,free list 應該按照記憶體地址的遞減順序來排序 * `(c)` best-fit 會選擇與請求記憶體區段匹配的最大 free list * `(d)` 安照 block size 遞增排序的 free list 上,使用 first-fit 和 best-fit 演算法等價 --- ### 測驗 `3` 考慮以下是利用 Linux epoll 系統呼叫開發的網頁伺服器,預期應該持續接受客戶端的連線 (port 55688) 並提供靜態內容。 > 透過網頁瀏覽器可取得 `Users Static Files` 字樣 ```cpp #include <errno.h> #include <fcntl.h> #include <netinet/in.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <sys/epoll.h> #include <unistd.h> #define MAX_EVENTS 4096 #define BACKLOG 1024 #define PORT 55688 #define CRLF "\r\n" #define DOUBLE_CRLF CRLF CRLF #define SOCKET_NON_BLOCKING(fd) \ int flags = fcntl(fd, F_GETFL, 0); \ if (flags == -1) \ abort(); \ flags |= O_NONBLOCK; \ if (fcntl(fd, F_SETFL, flags) == -1) \ abort(); #define EPOLL_CTL(efd, a, cfd, evs) \ if (epoll_ctl( \ efd, a, cfd, \ &(struct epoll_event){.events = evs, .data = {.fd = cfd}}) == -1) \ exit(EXIT_FAILURE); static const char *content = "HTTP/1.1 200 OK" CRLF "Content-Length: 18" CRLF "Content-Type: text/html" DOUBLE_CRLF "Users Static Files" DOUBLE_CRLF; static void do_use_fd(int epollfd, struct epoll_event *d) { if (d->events & MASK1) { char buf[512] = {0}; int n = read(d->data.fd, buf, 512); if (n > 0) { EPOLL_CTL(epollfd, EPOLL_CTL_MOD, d->data.fd, EPOLLOUT); } else { if (n == 0) { EPOLL_CTL(epollfd, EPOLL_CTL_DEL, d->data.fd, 0); close(d->data.fd); return; } } } else if (d->events & MASK2) { int n = write(d->data.fd, content, COUNT); if (n > 0) { EPOLL_CTL(epollfd, EPOLL_CTL_MOD, d->data.fd, EPOLLIN); } else { if (n == 0) { EPOLL_CTL(epollfd, EPOLL_CTL_DEL, d->data.fd, 0); close(d->data.fd); } if (errno != EAGAIN && errno != EWOULDBLOCK) close(d->data.fd); } } else { EPOLL_CTL(epollfd, EPOLL_CTL_DEL, d->data.fd, 0); close(d->data.fd); } } int main(void) { int listen_sock = socket(AF_INET, SOCK_STREAM | SOCK_NONBLOCK, IPPROTO_TCP); if (listen_sock == -1) exit(EXIT_FAILURE); struct sockaddr_in server_addr = {.sin_family = AF_INET, .sin_addr.s_addr = htonl(INADDR_ANY), .sin_port = htons(PORT)}; if ((bind(listen_sock, (struct sockaddr *) &server_addr, sizeof(server_addr))) != 0) exit(EXIT_FAILURE); if ((listen(listen_sock, BACKLOG)) != 0) exit(EXIT_FAILURE); struct linger linger = {.l_onoff = 1, .l_linger = 0}; if (setsockopt(listen_sock, SOL_SOCKET, SO_REUSEADDR, (const char *) &linger.l_onoff, sizeof(linger.l_onoff)) == -1) exit(EXIT_FAILURE); if (setsockopt(listen_sock, SOL_SOCKET, SO_LINGER, (const void *) &linger, sizeof(struct linger)) == -1) exit(EXIT_FAILURE); int epollfd = epoll_create1(0); if (epollfd == -1) exit(EXIT_FAILURE); struct epoll_event ev = {.events = EPOLLIN, .data.fd = listen_sock}; if (epoll_ctl(epollfd, EPOLL_CTL_ADD, listen_sock, &ev) == -1) exit(EXIT_FAILURE); struct epoll_event events[MAX_EVENTS]; for (;;) { int nfds = epoll_wait(epollfd, events, MAX_EVENTS, -1); if (nfds == -1) exit(EXIT_FAILURE); for (int n = 0; n < nfds; ++n) { if (events[n].data.fd == listen_sock) { ev.data.fd = accept(listen_sock, NULL, NULL); if (ev.data.fd == -1) exit(EXIT_FAILURE); SOCKET_NON_BLOCKING(ev.data.fd); ev.events = EPOLLIN; if (epoll_ctl(epollfd, EPOLL_CTL_ADD, ev.data.fd, &ev) == -1) exit(EXIT_FAILURE); } else do_use_fd(epollfd, &events[n]); } } return 0; } ``` 請補完程式碼,使得程式行為符合預期。 ==作答區== MASK1 = ? * `(a)` `EPOLLIN` * `(b)` `EPOLLOUT` MASK2 = ? * `(a)` `EPOLLIN` * `(b)` `EPOLLOUT` COUNT = ? * `(a)` 16 * `(b)` 26 * `(c)` 36 * `(d)` 46 * `(e)` 56 * `(f)` 66 * `(g)` 86