Question

string s1 = Encoding.ASCII.GetString(buf2);

我看到类型字符串只限制8个字节（8个字符）。大家可以告诉我如何接收超过8个字节的方式？详情：

服务器

        byte[]  buf = Encoding.ASCII.GetBytes(fileSize.ToString());
        MemoryStream ms = new MemoryStream();    
        ms.Write(buf,0,buf.Length);
        ms.Seek(0, SeekOrigin.Begin);
        byte[] buf2 = new byte[32*1024];
        ms.Read(buf2,0,buf2.Length);
        FileStream fs = File.OpenRead(savefilename);
        MessageBox.Show("1 " + buf2.Length +" "+ fileSize+" "+ms.Length);
        int ns = sendSocket.Send(buf2, sizeof(ulong), SocketFlags.None);

客户端

            byte[] buf = new byte[32 * 1024];
            int nr = listenSocket2.Receive(buf, sizeof(ulong), SocketFlags.None);
            MemoryStream ms = new MemoryStream();
            ms.Write(buf, 0, buf.Length);
            ms.Seek(0, SeekOrigin.Begin);
            byte[] buf2 = new byte[32 * 1024];
            ms.Read(buf2, 0, buf2.Length);

            string s1 = Encoding.ASCII.GetString(buf2);
            MessageBox.Show("" + s1+" "+ms.Length);
            ulong fileSize = UInt64.Parse(s1);

s1不超过8个字符

Answer 1

以下一行

int nr = listenSocket2.Receive(buf, sizeof(ulong), SocketFlags.None);

从流中读取前8个字节。这意味着包含文件的 size 。您应该发出另一个Receive来接收实际的文件内容。

大小类型字符串

1 个答案: