如何组合两个字节来获取c＃中的原始值 [英] How to combine two bytes to get the original value in c#
问题描述
美好的一天，
我需要你的帮助和支持，如何组合两个字节（每个字节是8位宽），以获得原始值。我正在GUI上显示从我的微控制器发送的值（从0到500）。我可以成功显示0到255之间的值，因为这只需要发送一个字节。但是，从256到500发送值需要发送两个字节。我遇到的问题是我无法重新组合收到的两个字节以获得原始值。以下是我的代码行：
Good day to all,
Please I need your help and support on how to combine two bytes (each byte is 8 bits wide) in order to get the original value. I am displaying values(from 0 to 500)sent from my microcontroller on the GUI.I can successfully display values from 0 to 255 as this requires just sending a byte. However sending values from 256 to 500 requires sending two bytes. The problem I am having is that I was unable to re-combine the received two bytes in order to get the original value. Below are my lines of code:
int main( void ) // this is the main function inside the microcontroller
02 {
03 sei();
04 USI_TWI_Master_Initialise();
05 Spi_Master_Init();
06
07
08 while(1) // I am using this loop to send 500(111110100)
09 {
10
11
12 Transmitt_Receive(244);//this function sends LOW BYTE of 500(11110100)
13
14 _delay_ms(1000);
15
16 Transmitt_Receive(1); // this function sends HIGH BYTE of 500(00000001)
17 _delay_ms(1000);
18
19
20
21 }
22 }
private void rxtemp_CheckedChanged(object sender, EventArgs e)
02 {
03 UInt32 numBytesRead = 0;
04 UInt32 numBytesToRead =1 ;
05 Int32 value;
06
07 byte[] readData = new byte[5];
08 byte[] val = new byte[5];
09
10 Thread th = new Thread(() =>
11 {
12 while (true)
13 {
14 ftStatus = myFtdiDevice.Read(readData, numBytesToRead, ref numBytesRead);
15 val[0] = readData[0];
16 ftStatus = myFtdiDevice.Read(readData, numBytesToRead, ref numBytesRead);
17 val[1] = readData[0];
18
19 value = (val[0] << 8) | val[1];
20
21 label3.Invoke(new Action(() =>
22 {
23 label3.Text = value + "ºC";
24
25
26 }));
27
28
29
30
31 }
32
33
34 });
35 th.IsBackground = true;
36 th.Start();
37 }
When I combined the two bytes using the above arrangements, instead of getting 500 my GUI displayed 62708.I got the same result when I used BitConverter.
<pre lang="c#">
value3 = BitConverter.ToInt32(val,0);
如果有人能告诉我如何解决这个错误，我会很高兴。谢谢你的支持。最好的问候。
I would be very glad if somebody could put me through on how to resolve this error. Thank you for the support. Best regards.
推荐答案
两个字节是16位。请尝试BitConverter.ToUInt16。此外，一次读取两个字节而不是一个。网络字节是big-endian，你的系统是little-endian。您现在将它们作为小端处理，混合字节顺序，导致错误的值。
一些关于这些endians的额外信息（实际上来自Gulliver's Travels）：
http://en.wikipedia.org/wiki/Endianness [ ^ ]
祝你好运！
Two bytes is 16 bit. TryBitConverter.ToUInt16instead. Also, read the two bytes at once instead one at a time. Network bytes are big-endian and your system is little-endian. You now handle them as little endian, mixing up the byte-order, resulting in a faulthy value.
Some nice extra info about these endians (actually from Gulliver’s Travels):
http://en.wikipedia.org/wiki/Endianness[^]
Good luck!
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