How to get little endian data from big endian in c# using bitConverter.ToInt32 method?

In modern-day Linq the one-liner and easiest to understand version would be:

int number = BitConverter.ToInt32(waveData.Skip(286).Take(4).Reverse().ToArray(), 0);

You could also...

byte[] tempForTimestamp = new byte[4];
Array.Copy(waveData, 287, tempForTimestamp, 0, 4);
Array.Reverse(tempForTimestamp);
int number = BitConverter.ToInt32(tempForTimestamp);

:)

If you know the data is big-endian, perhaps just do it manually:

int value = (buffer[i++] << 24) | (buffer[i++] << 16)
          | (buffer[i++] << 8) | buffer[i++];

this will work reliably on any CPU, too. Note i is your current offset into the buffer.

Another approach would be to shuffle the array:

byte tmp = buffer[i+3];
buffer[i+3] = buffer[i];
buffer[i] = tmp;
tmp = buffer[i+2];
buffer[i+2] = buffer[i+1];
buffer[i+1] = tmp;
int value = BitConverter.ToInt32(buffer, i);
i += 4;

I find the first immensely more readable, and there are no branches / complex code, so it should work pretty fast too. The second could also run into problems on some platforms (where the CPU is already running big-endian).

Here you go

public static int SwapEndianness(int value)
{
    var b1 = (value >> 0) & 0xff;
    var b2 = (value >> 8) & 0xff;
    var b3 = (value >> 16) & 0xff;
    var b4 = (value >> 24) & 0xff;

    return b1 << 24 | b2 << 16 | b3 << 8 | b4 << 0;
}