小编Syl*_*lar的帖子

我刚开始用C学习网络编程.我做了一些测试,但是我遇到了错误.

我有一个客户:

client.c

#include <string.h>
#include <netdb.h>
#include <stdio.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <errno.h>

int main(void)
{
    struct addrinfo hints, *res;
    int sock;

    memset(&hints, 0, sizeof hints);
    hints.ai_family = AF_UNSPEC;
    hints.ai_socktype = SOCK_STREAM;

    getaddrinfo("localhost", "5996", &hints, &res);
    sock = socket(res->ai_family, res->ai_socktype, res->ai_protocol);
    connect(sock, res->ai_addr, res->ai_addrlen);
    char data[64];
    int len = 13;
    int br = recv(sock, data, len, 0);
    printf("%s\n%s\n%d\n", strerror(errno), data, br);
    return 0;
}

和服务器:

server.c

#include <string.h>
#include <netdb.h>
#include <stdio.h>
#include <sys/types.h>
#include <sys/socket.h> …

因此，我有一个分为两个三角形的矩形，并且有一个点：

我知道矩形的四个顶点（A，B，C和D）的坐标，也知道P的坐标。我需要知道的是两个三角形中的哪个是点。

我认为这是一件非常简单的事情，但是显然我自己无法弄清楚。

有什么帮助吗？

我正在尝试将秒数转换为日期.这是我目前拥有的(在PHP中):

function secondsToTime($inputSeconds) {
    $secondsInAMinute = 60;
    $secondsInAnHour  = 60 * $secondsInAMinute;
    $secondsInADay    = 24 * $secondsInAnHour;
    $secondsInAMonth = 30 * $secondsInADay;
    $secondsInAYear = 12 * $secondsInAMonth;

    $years = floor($inputSeconds / $secondsInAYear);

    $monthSeconds = $inputSeconds % $secondsInAYear;
    $months = floor($monthSeconds / $secondsInAMonth);

    $daySeconds = $monthSeconds % $secondsInAMonth;
    $days = floor($daySeconds / $secondsInADay);

    $hourSeconds = $daySeconds % $secondsInADay;
    $hours = floor($hourSeconds / $secondsInAnHour);

    $minuteSeconds = $hourSeconds % $secondsInAnHour;
    $minutes = floor($minuteSeconds / $secondsInAMinute);

    $remainingSeconds = $minuteSeconds % $secondsInAMinute;
    $seconds = ceil($remainingSeconds); …