what is actually stored after converting audio to byte array in java?

Question

Actually i am implementing audio steganography using the LSB replacement algorithm. I am converting the audio to byte array using the following code:

    File src = new File("C:\\test.wav");
    AudioInputStream ais = AudioSystem.getAudioInputStream(src);
    byte[] data = new byte[ais.available()];
    int n = ais.read(data);

Now the thing is that this audio file is a 16bit sample length, dual channel and LittleEndian; i checked this using the "isBigEndian()" function. Thus, i want to know how are the samples stored in the byte array so that i know which bits to replace.

Answer 1

How are you converting audio to a byte array?

What is actually stored is an array of samples. If you have a clean single tone of a certain frequency you would get an array that follows a sine curve. Here is an example of what you will see for a perfect sine wave:

127,130,133,136,139,142,145,148,151,154,157,160,163,166,169,172,175,178,181,184,186,189,192,194,197,200,202,205,207,209,212,214,216,218,221,223,225,227,229,230,232,234,235,237,239,240,241,243,244,245,246,247,248,249,250,250,251,252,252,253,253,253,253,253,254,253,253,253,253,253,252,252,251,250,250,249,248,247,246,245,244,243,241,240,239,237,235,234,232,230,229,227,225,223,221,218,216,214,212,209,207,205,202,200,197,194,192,189,186,184,181,178,175,172,169,166,163,160,157,154,151,148,145,142,139,136,133,130,127,123,120,117,114,111,108,105,102,99,96,93,90,87,84,81,78,75,72,69,67,64,61,59,56,53,51,48,46,44,41,39,37,35,32,30,28,26,24,23,21,19,18,16,14,13,12,10,9,8,7,6,5,4,3,3,2,1,1,0,0,0,0,0,0,0,0,0,0,0,1,1,2,3,3,4,5,6,7,8,9,10,12,13,14,16,18,19,21,23,24,26,28,30,32,35,37,39,41,44,46,48,51,53,56,59,61,64,67,69,72,75,78,81,84,87,90,93,96,99,102,105,108,111,114,117,120,123

If you have a 16-bit stream you should use a short[] to store it so you don't need to worry about big/little endian. If you choose to use a byte[] it will be twice the size as the number of audio samples and two consecutive bytes would need to be combined to represent one sample.