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Home/ Questions/Q 8919711
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T06:05:41+00:00 2026-06-15T06:05:41+00:00

I’m confused. I needed to record sound from MIC in Android so I used

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I’m confused. I needed to record sound from MIC in Android so I used the following code:

recorder = new AudioRecord(AudioSource.MIC, 44100,
                AudioFormat.CHANNEL_IN_MONO,
                AudioFormat.ENCODING_PCM_16BIT, N);
buffer = new byte[N];
//...
recorder.read(buffer, 0, N);

As we know, a byte array can store values between -128 to +128 while a 16Bit sound needs a lot more storage(e.g. short and int) but surprisingly Java and Android have a record method which saves recorded data to a byte array.

How that can be possible? What am I missing?

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  1. Editorial Team
    Editorial Team
    2026-06-15T06:05:42+00:00Added an answer on June 15, 2026 at 6:05 am

    You are thinking of byte as a shot integer. It is just 8 bits. You need to store 1000111011100000 (16 bits)? First byte is 10001110, second byte is 11100000. That you can interpret these bits as numbers is not relevant here. In a more general way, byte[] is usually how you deal with binary “raw data” (let it be audio streams or encrypted content or anything else that you treat like a stream of bits).

    If you have n “words” of 16 bits then you will need 2n bytes to store it. Byte 0 will be lower (or higher) part of word 0, byte 1 will be the rest of word 0, byte 0 will be lower (or higher) part of word 1…

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