Endianness on Android when using AudioRecord with 16-bit PCM and reading into a byte array

Question

I am currently examining an Android application that uses AudioRecord to record audio in 16-bit PCM format:

byte [] buffer = new byte[1600];
audioRecord.read(buffer, 0, 1600);

It stores the recorded audio into buffer.The documentation of this read function describes that this function should only be used with 8-bit PCM. However, the Android application uses it with 16-bit PCM (and it seems to work without issues; another overloaded read variant using a byte array also mentions that the use of 16-bit PCM with this method is possible, but deprecated).

Now I am unsure if each sample (consisting of 2 bytes) is stored in little or in big endian format. The documentation section about the audio encoding says that using a ByteBuffer instead of byte array results in native endian (instead of big endian).

I suspect that a short is stored in big endian format but I can not find evidence for this.

Answer 1

I have the same situation in my recent project.

Method

public int read (byte[] audioData, 
            int offsetInBytes, 
            int sizeInBytes)

actually calls

public int read (byte[] audioData, 
            int offsetInBytes, 
            int sizeInBytes, 
            int readMode)

with AudioRecord.READ_BLOCKING as the 4th parameter. Since 16-bit PCM is deprecated, I think it's still ok to use this method just not recommended.

The read method finally calls native method native_read_in_byte_array to fill the audio buffer. Android is natively little-endian so native_read_in_byte_array stores audio data in little endian in C/C++ layer and then send audio data back to Java layer through JNI.

I did a quick test to find that byte buffer passing through JNI won't change the order in which it stores its bytes.

Basically you get your Java byte[] in the same order as in the C/C++ jbyteArray. So a short in native is still stored in little endian in Java layer.

That's all I can reason from my test. Hope it helps and let me know if something is wrong in there.