Convert signed int (2bytes, 16 bits) in double format. With Java

Question

I've got a problem. In Java I need to read samples from a wav file. The file format is: wav, PCM_SIGNED, signed int of 2bytes = 16bits, little endian... The object reads the audio samples in BYTES and I need to convert this two bytes in one double value. I tried to use this formula but it's not completely correct:

mono = (double)((audioBytes[k] & 0xFF) | (audioBytes[k + 1] << 8));

Comparing the results with Matlab I always notice differences between the real value in Matlab and the converted one in Java. Can anybody help me please? Thank you, Dave

Answer 1

You haven't given us enough information to know why you're getting different results in Matlab and Java. Usually you scale the short channel data [-32768..32767] to a double in the range [-1..1] which it looks like you are attempting to do. Your java result: -3.0517578125E-5 is correct for the short value -1: -1/32768. I don't know why your Matlab result is different. You haven't shown us how you are arriving at your Matlab results.

If you have a large sequence of bytes (which I'm guessing you do), and you don't want to worry about BIG-ENDIAN vs LITTLE-ENDIAN or shifting bits and bytes, let java take care of it for you:

import java.nio.*;
...
ByteBuffer buf = ByteBuffer.wrap(audioBytes);
buf.order(ByteOrder.LITTLE_ENDIAN);

while (buf.remaining() >= 2) {
    short s = buf.getShort();
    double mono = (double) s;
    double mono_norm = mono / 32768.0;
    ...
}

ByteBuffer.getShort() reads the next two bytes of the buffer, takes care of the Little-Endian ordering, converts the bytes to a short, and positions itself for the next getXXX() call.

Answer 2

This is the correct way:

double sampleValue = (double)(( bytes[0]<<8 ) | ( bytes[1]&0x00FF ));

(Change indices to swap little/big)

Answer 3

Cant you just do most_significant byte * 256 + least_significant_byte and then cast to double?