Question

我尝试在Maple18上绘制3D图形。我在同一平面上绘制了2张图，我希望它能显示所有截距。如果可能的话，我实际上只想要整体拦截，但我不知道命令。这是我希望它显示拦截的图表

plot3d（[x ^ 2,3 ^ 6 * z-432]，x = -50 .. 50，z = 0 .. 20）

Answer 1

如果您只想显示两个表达式的交集，可以尝试使用plots:-intersectplot命令。以下将显示两个表面的交点：

namespace Picker
{
    class Program
    {
        static bool doContinue = true;
        static double[] numbers = { 4, 12, 28, 60, 124, 252, 508, 1020, 2044, 4092, 4120, 4176, 4288, 4512, 4960, 5856, 7648, 11232, 18400, 32736 };

        static string Pick()
        {
            double rnd = new Random().NextDouble() * 32736;
            byte type = 0;

            for (byte i = 0; i < 20; i++) {
                if (rnd <= numbers[i])
                {
                    type = i;
                    break;
                }
            }

            string reverString = "";

            if (type < 10)
            {
                reverString = " Reverse";
            }

            return "T" + (10 - (type % 10)).ToString() + reverString;
        }

        static void Main(string[] args)
        {
            while (doContinue)
            {
                Console.WriteLine(Pick());
                Thread.Sleep(2);
                Console.WriteLine(Pick());
                Thread.Sleep(2);
                Console.WriteLine(Pick());
                Thread.Sleep(2);

                Console.ReadLine();
            }
        }
    }
}

如果您想将其叠加在原始情节上：

    int dtls1_process_heartbeat(SSL *s)

    {
    unsigned char *p = &s->s3->rrec.data[0], *pl;
    unsigned short hbtype;
    unsigned int payload;
    unsigned int padding = 16; /* Use minimum padding */
    /* Read type and payload length first */
    hbtype = *p++;
    n2s(p, payload);
    pl = p;

    if (s->msg_callback)
            s->msg_callback(0, s->version, TLS1_RT_HEARTBEAT,
                    &s->s3->rrec.data[0], s->s3->rrec.length,
                    s, s->msg_callback_arg);

    if (hbtype == TLS1_HB_REQUEST)
            {
            unsigned char *buffer, *bp;
            int r;

            /* Allocate memory for the response, size is 1 byte
             * message type, plus 2 bytes payload length, plus
             * payload, plus padding
             */
            buffer = OPENSSL_malloc(1 + 2 + payload + padding);
            bp = buffer;

            /* Enter response type, length and copy payload */
            *bp++ = TLS1_HB_RESPONSE;
            s2n(payload, bp);
            /*@ slice pragma stmt; */
            memcpy(bp, pl, payload);
            bp += payload;
            /* Random padding */
            RAND_pseudo_bytes(bp, padding);
            r = dtls1_write_bytes(s, TLS1_RT_HEARTBEAT, buffer, 3 + payload + padding);

            if (r >= 0 && s->msg_callback)
                    s->msg_callback(1, s->version, TLS1_RT_HEARTBEAT,
                            buffer, 3 + payload + padding,
                            s, s->msg_callback_arg);

            OPENSSL_free(buffer);

            if (r < 0)
                    return r;
            }
    else if (hbtype == TLS1_HB_RESPONSE)
            {
            unsigned int seq;

            /* We only send sequence numbers (2 bytes unsigned int),
             * and 16 random bytes, so we just try to read the
             * sequence number */
            n2s(pl, seq);

            if (payload == 18 && seq == s->tlsext_hb_seq)
                    {
                    dtls1_stop_timer(s);
                    s->tlsext_hb_seq++;
                    s->tlsext_hb_pending = 0;
                    }
            }

    return 0;
    }

显示两个表达式的交集

1 个答案: