csapp ch11.04 练习题

这个题目的解答的核心是知道这两个数据结构可以互换

下面的输出是

    printf("uint16_t: %ld\n", sizeof(uint16_t));
    printf("in_addr: %ld\n", sizeof(struct in_addr));

2+4+8恰好14
例子代码

#include "csapp.h"

int main (int argc, char **argv) {
    struct addrinfo *p, *listp, hints;
    char buf[MAXLINE];
    int rc, flags;

    if (argc != 2) {
        fprintf(stderr, "usage: %s \n", argv[0]);
        exit(0);
    }
    memset(&hints, 0, sizeof(struct addrinfo));
    hints.ai_family = AF_INET;
    hints.ai_socktype = SOCK_STREAM;
    if ((rc = getaddrinfo(argv[1], NULL, &hints, &listp))!= 0) {
        fprintf(stderr, "getaddrinfo error: %s\n", gai_strerror(rc));
        exit(1);
    }
    flags = NI_NUMERICHOST;
    for (p = listp; p; p = p->ai_next) {
        Getnameinfo(p->ai_addr, p->ai_addrlen, buf, MAXLINE, NULL, 0, flags);
        printf("%s\n", buf);
    }
    Freeaddrinfo(listp);
    exit(0);
}

我的实现代码

#include "csapp.h"
#include 
int main (int argc, char **argv) {
    struct addrinfo *p, *listp, hints;
    char buf[MAXLINE];
    int rc, flags;

    if (argc != 2) {
        fprintf(stderr, "usage: %s \n", argv[0]);
        exit(0);
    }
    memset(&hints, 0, sizeof(struct addrinfo));
    hints.ai_family = AF_INET;
    hints.ai_socktype = SOCK_STREAM;
    if ((rc = getaddrinfo(argv[1], NULL, &hints, &listp))!= 0) {
        fprintf(stderr, "getaddrinfo error: %s\n", gai_strerror(rc));
        exit(1);
    }
    flags = NI_NUMERICHOST;
    for (p = listp; p; p = p->ai_next) {
        struct sockaddr_in* sin = (struct sockaddr_in*)p->ai_addr;
        struct in_addr src = sin->sin_addr;
        char dst[20];
        inet_ntop(AF_INET, (void*)&src, dst, INET_ADDRSTRLEN);
        printf("%s\n", dst);
    }
    Freeaddrinfo(listp);
    exit(0);
}

不知道为什么和前面那个顺序是反的
答案

解这个问题的时候搜了一下百度，当时是想弄清楚sockaddr是怎么回事，结果

当我知道要转换后，这个问题怎么解基本思路有了，只是不知道会不会有什么意外，比如大小端什么的