在 C 中将字节转换为 Int/uint [英] Convert Bytes to Int / uint in C

问题描述

我有一个无符号字符数组[248]；充满字节.像 2F AF FF 00 EB AB CD EF .....这个数组是我的字节流，我将来自 UART (RS232) 的数据作为缓冲区存储.

现在我想将字节转换回我的 uint16 和 int32.

在 C# 中，我使用 BitConverter 类来执行此操作.例如:

byte[] Array = { 0A, AB, CD, 25 };int myint1 = BitConverter.ToInt32(bytes, 0);int myint2 = BitConverter.ToInt32(bytes, 4);int myint3 = BitConverter.ToInt32(bytes, 8);int myint4 = BitConverter.ToInt32(bytes, 12);//...在此处输入代码Console.WriteLine("int: {0}", myint1);//输出数据...

C 中有类似的函数吗?(没有 .net ，我使用 KEIL 编译器，因为代码在微控制器上运行)

问候山姆

解决方案

是的.假设您的字节在:

uint8_t bytes[N] = {/* 随便 */};

我们知道，一个 16 位整数只是两个 8 位整数连接在一起，即一个是 256 的倍数或移位 8:

uint16_t 十六[N/2];对于 (i = 0; i 

同样适用于 32 位:

uint32_t三十二[N/4];对于 (i = 0; i 

<小时>

如果字节是大端读取的，当然你要颠倒顺序:

bytes[i+1] |(uint16_t)bytes[i] <<8

和

bytes[i+3] |(uint32_t)bytes[i+2] <<8|(uint32_t)bytes[i+1] <<16 |(uint32_t)bytes[i] <<24

请注意，存储整数的字节序与运行架构的字节序之间存在差异.这个答案中提到的字节序是存储的整数，即 bytes 的内容.由于移位时会考虑字节序，因此解决方案与正在运行的架构的字节序无关.>

I have a unsigned char array[248]; filled with bytes. Like 2F AF FF 00 EB AB CD EF ..... This Array is my Byte Stream which I store my Data from the UART (RS232) as a Buffer.

Now I want to convert the bytes back to my uint16's and int32's.

In C# I used the BitConverter Class to do this. e.g:

byte[] Array = { 0A, AB, CD, 25 };
int myint1 = BitConverter.ToInt32(bytes, 0);
int myint2 = BitConverter.ToInt32(bytes, 4);
int myint3 = BitConverter.ToInt32(bytes, 8);
int myint4 = BitConverter.ToInt32(bytes, 12);
//...
enter code here
Console.WriteLine("int: {0}", myint1); //output Data...

Is there a similiar Function in C ? (no .net , I use the KEIL compiler because code is running on a microcontroller)

With Regards Sam

解决方案

Yes there is. Assume your bytes are in:

uint8_t bytes[N] = { /* whatever */ };

We know that, a 16 bit integer is just two 8 bit integers concatenated, i.e. one has a multiple of 256 or alternatively is shifted by 8:

uint16_t sixteen[N/2];

for (i = 0; i < N; i += 2)
    sixteen[i/2] = bytes[i] | (uint16_t)bytes[i+1] << 8;
             // assuming you have read your bytes little-endian

Similarly for 32 bits:

uint32_t thirty_two[N/4];

for (i = 0; i < N; i += 4)
    thirty_two[i/4] = bytes[i] | (uint32_t)bytes[i+1] << 8
        | (uint32_t)bytes[i+2] << 16 | (uint32_t)bytes[i+3] << 24;
             // same assumption

---

If the bytes are read big-endian, of course you reverse the order:

bytes[i+1] | (uint16_t)bytes[i] << 8

and

bytes[i+3] | (uint32_t)bytes[i+2] << 8
    | (uint32_t)bytes[i+1] << 16 | (uint32_t)bytes[i] << 24

Note that there's a difference between the endian-ness in the stored integer and the endian-ness of the running architecture. The endian-ness referred to in this answer is of the stored integer, i.e., the contents of bytes. The solutions are independent of the endian-ness of the running architecture since endian-ness is taken care of when shifting.

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