3转换成字节有符号整数在C＃ [英] Converting 3 bytes into signed integer in C#

问题描述

我想3个字节转换为C＃符号整数（大端）。

I'm trying to convert 3 bytes to signed integer (Big-endian) in C#.

我试图用 BitConverter .ToInt32 方法，但我的问题是什么样的价值应该有拉特字节。

I've tried to use BitConverter.ToInt32 method, but my problem is what value should have the lats byte.

任何人都可以建议我，我怎么能做到这一点的方式不同？

Can anybody suggest me how can I do it in different way?

我还需要转换5（或6或7）个字节来签订长，是没有任何一般规则该怎么办呢？

I also need to convert 5 (or 6 or 7) bytes to signed long, is there any general rule how to do it?

感谢您事先的任何帮助。

Thanks in advance for any help.

推荐答案

作为最后的手段，你可以随时切换+添加自己：

As a last resort you could always shift+add yourself:

byte b1, b2, b3;

int r = b1 << 16 + b2 << 8 + b3;

只要将B1 / B2 / B3，直到你想要的结果。

Just swap b1/b2/b3 until you have the desired result.

在第二个想法，这绝不会产生负值。结果
你想要什么结果当MSB> = 0x80的？

On second thought, this will never produce negative values.
What result do you want when the msb >= 0x80 ?

2部分，蛮力符号扩展：

Part 2, brute force sign extension:

    private static int Bytes2Int(byte b1, byte b2, byte b3)
    {
        int r = 0;
        byte b0 = 0xff;

        if ((b1 & 0x80) != 0) r |= b0 << 24;
        r |= b1 << 16;
        r |= b2 << 8;
        r |= b3;
        return r;
    }

我已经测试这b $ b

I've tested this with:

      byte[] bytes = BitConverter.GetBytes(p);
      int r = Bytes2Int(bytes[2], bytes[1], bytes[0]);
      Console.WriteLine("{0} == {1}", p, r);

几 P 。

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