BitConverter.ToInt32 to convert 2 bytes

Question

I am using BitConverter.ToInt32 to convert a Byte array into int.

I have only two bytes [0][26], but the function needs 4 bytes, so I have to add two 0 bytes to the

Answer 1

You should probably do (int)BitConverter.ToInt16(..) instead. ToInt16 is made to read two bytes into a short. Then you simply convert that to an int with the cast.

Answer 2

Array.Copy. Here is some code:

byte[] arr = new byte[] { 0x12, 0x34 };
byte[] done = new byte[4];
Array.Copy(arr, 0, done, 2, 2); // http://msdn.microsoft.com/en-us/library/z50k9bft.aspx
int myInt = BitConverter.ToInt32(done); // 0x00000026

However, a call to `BitConverter.ToInt16(byte[]) seems like a better idea, then just save it to an int:

int myInt = BitConverter.ToInt16(...);

Keep in mind endianess however. On little endian machines, { 0x00 0x02 } is actually 512, not 2 (0x0002 is still 2, regardless of endianness).

Answer 3

You should call `BitConverter.ToInt16, which only reads two bytes.

short is implicitly convertible to int.