java - 字节数组到短数组并在java中再次返回

Question

我在获取存储在字节数组中的音频数据、将其转换为大端短数组、对其进行编码、然后将其更改回字节数组时遇到了一些问题。这就是我所拥有的。原始音频数据存储在 audioBytes2 中。我使用相同的格式进行解码，但在 cos 函数上加减号。不幸的是，更改字节和短数据类型是不可协商的。

    short[] audioData = null;
    int nlengthInSamples = audioBytes2.length / 2;
    audioData = new short[nlengthInSamples];

    for (int i = 0; i < nlengthInSamples; i++) {
       short MSB = (short) audioBytes2[2*i+1];
       short LSB = (short) audioBytes2[2*i];
       audioData[i] = (short) (MSB << 8 | (255 & LSB));
    }

    int i = 0;
    while (i < audioData.length) {
        audioData[i] = (short)(audioData[i] + (short)5*Math.cos(2*Math.PI*i/(((Number)EncodeBox.getValue()).intValue())));
        i++;
    }

    short x = 0;
    i = 0;
    while (i < audioData.length) {
        x = audioData[i];
        audioBytes2[2*i+1] = (byte)(x >>> 0);
        audioBytes2[2*i] = (byte)(x >>> 8);
        i++;
    }

我已经尽我所能来完成这项工作，但我最接近的方法是让它在所有其他编码/解码中工作，我不知道为什么。感谢您的帮助。

Answer 1

我也建议你试试 ByteBuffer。

byte[] bytes = {};
short[] shorts = new short[bytes.length/2];
// to turn bytes to shorts as either big endian or little endian. 
ByteBuffer.wrap(bytes).order(ByteOrder.LITTLE_ENDIAN).asShortBuffer().get(shorts);

// to turn shorts back to bytes.
byte[] bytes2 = new byte[shortsA.length * 2];
ByteBuffer.wrap(bytes2).order(ByteOrder.LITTLE_ENDIAN).asShortBuffer().put(shortsA);