我很难从C服务器传递一个简短值。我在另一端收到的是它看起来像是'垃圾',我似乎无法将其转化为有用的东西。C服务器到Java客户端套接字读取传递的值很短
因素我认为是影响它
- 二进制补
- 潜在ASCII或UTF-8 enconding
与C语言代码开始:
// Read 10 bit value into short
unsigned short ret = decideActionCall(buffer);
unsigned char lsb = (unsigned char) ret;
unsigned char msb = (unsigned char) (ret >> 8);
printf("msb,lsb = %02x,%02x\n",msb, lsb);
/**
char retString [4];
retString[0] = msb;
retString[1] = lsb;
retString[2] = '\n';
retString[3] = '\0';
*/
unsigned short retString1[1];
retString1[0] = ret;
printf("msb,lsb = %d,%d\n",retString[0],retString[1]);
printf("String to send: <%s>\n",retString1);
//if (write(newsockfd, retString, 3) < 0) {
// error("Error sending response to the server");
//}
if (write(newsockfd, &retString1, 2) < 0) {
error("Error sending response to the server");
}
if (write(newsockfd, "\r",1) < 0) {
error("Error sending response to the server");
}
和终端输出是
START-----------------------------
0 0
136 0x88
ADC3 = 136
END-----------------------------
decideActionCall() ret value: 136
msb,lsb = 00,88
msb,lsb = 0,136
String to send: <�>
所以'ret'的返回值是介于0和1023(0x3FF)之间的值。这10位需要从C服务器传输到客户端。我尝试过两种实现方式,我将它们发送为8位字符(retString)或一个16位短(retString1)。 Sizeof(char) == 1
和sizeof(short) == 2
。
的Java代码(有各种尝试解析它):
BufferedReader bufIn = new BufferedReader(new InputStreamReader(in));
String response = bufIn.readLine();
char [] hello = response.toCharArray();
for (char a : hello) {
short shortA = (short) (a & 0x3FF);
System.out.println(a + " = " + (int)a + " = " + shortA);
//System.out.println((int)a);
//System.out.println(Integer.valueOf(a+"",10));
}
byte test[] = response.getBytes("UTF-8");
StringBuilder sb = new StringBuilder();
for (byte b : test) {
sb.append(String.format("%02X ", b));
}
System.out.println(sb.toString());
byte b05 = 0x00;
byte b06 = 0x00; // MSB, positive as < 0x80
int i = 0;
for (byte a : test) {
//System.out.println(short1);
System.out.println("byte: " + a);
//System.out.println(Integer.valueOf(a+"",10));
if (i == 0) {
b05 = a;
} else {
b06 = a;
}
i++;
}
System.out.println("i = " + i);
byte[] byteTabDay = new byte[2];
byteTabDay[0] = b05;
byteTabDay[1] = b06;
BigInteger temp = new BigInteger(test);
System.out.println(temp);
System.out.println(temp.intValue());
System.out.println(temp.shortValue() + 65536);
BigInteger temp2 = new BigInteger(byteTabDay);
System.out.println(temp2);
System.out.println(temp2.shortValue() + 65536);
Java示例输出(Android的logcat的):
I/System.out﹕ response: ���
I/System.out﹕ here: � 65533
I/System.out﹕ ascii: 65485 65581
I/System.out﹕ � = 65533 = 1021
I/System.out﹕ �� = 0 = 0
I/System.out﹕ EF BF BD 00
I/System.out﹕ byte: -17
I/System.out﹕ byte: -65
I/System.out﹕ byte: -67
I/System.out﹕ byte: 0
I/System.out﹕ i = 4
I/System.out﹕ -272646912
I/System.out﹕ -272646912
I/System.out﹕ 48384
I/System.out﹕ -4352
I/System.out﹕ 61184
问题:
- 我应该如何正确读取套接字?
- 如何正确读取10位? (我已经尝试用0x3FF掩盖字符)
- 是两个补码可能导致问题吗?
谢谢
如果你想要一个16位整数,不要使用'short',因为short的大小是实现定义的。改为使用'uint16_t' - 它被定义为16位。您还想按网络顺序读取和写入多字节整数。也尝试用十六进制打印你的Java输出。 – 2015-04-03 23:06:02
谢谢你的回应。我将这些值更改为uint16_t和unint8_t(如果可用)。我还使用1和0给出的链接将其更改为网络字节顺序。我也打印了十六进制,但仍然混乱。我实施了EJP建议的更改,似乎已经奏效。 – koala123 2015-04-03 23:33:37
此外,请考虑使用协议缓冲区而不是手动推送二进制数据。 – chrylis 2015-04-04 00:04:13