c# – 从.WAV文件中读取24位样本

private List<float> Read32BitSamples(FileStream stream,int sampleStartIndex,int sampleEndIndex)
{
    var samples = new List<float>();
    var bytes = ReadChannelBytes(stream,Channels.Left,sampleStartIndex,sampleEndIndex); // Reads bytes of a single channel.

    if (audioFormat == WavFormat.PCM)  // audioFormat determines whether to process sample bytes as PCM or floating point.
    {
        for (var i = 0; i < bytes.Length / 4; i++)
        {
            samples.Add(BitConverter.ToInt32(bytes,i * 4) / 2147483648f);
        }
    }
    else
    {
        for (var i = 0; i < bytes.Length / 4; i++)
        {
            samples.Add(BitConverter.ToSingle(bytes,i * 4));
        }
    }

    return samples;
}

读取(和存储)24位样本非常简单.现在,正如你所说的那样,框架中不存在3字节整数类型,这意味着你有两种选择;要么创建自己的类型,要么通过在样本的字节数组的开头插入一个空字节(0)来填充24位样本,从而使它们成为32位样本(这样你就可以使用int来存储/操纵它们).

我将解释并演示如何做到以后(这也是我认为更简单的方法).

首先,我们必须看看如何将24位样本存储在int中,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ MSB ~ ~ 2^ndMSB ~ ~ 2^ndLSB ~ ~ LSB ~ ~

24-bit sample: 11001101 01101001 01011100 00000000

32-bit sample: 11001101 01101001 01011100 00101001

MSB =最高有效字节,LSB =无效有效字节.

正如您所看到的,24位样本的LSB为0,因此您只需要声明一个包含4个元素的byte [],然后将样本的3个字节读入数组(从元素1开始),这样您的数组如下所示(有效地向左移位8位),

myArray[0]: 00000000

myArray[1]: 01011100

myArray[2]: 01101001

myArray[3]: 11001101

一旦你的字节数组已满,就可以将它传递给BitConverter.ToInt32(myArray,0);,然后你需要将样本向右移动8个位置以获得样本的正确的24位字节表示(从-8388608到8388608);然后除以8388608将其作为浮点值.

所以,把所有这些放在一起你应该最终得到这样的东西,

注意,我编写了以下代码,意图是“易于遵循”,因此这不是最高效的方法,对于更快的解决方案,请参阅下面的代码.

private List<float> Read24BitSamples(FileStream stream,int startIndex,int endIndex)
{
    var samples = new List<float>();
    var bytes = ReadChannelBytes(stream,startIndex,endIndex);
    var temp = new List<byte>();
    var paddedBytes = new byte[bytes.Length / 3 * 4];

    // Right align our samples to 32-bit (effectively bit shifting 8 places to the left).
    for (var i = 0; i < bytes.Length; i += 3)
    {
        temp.Add(0);            // LSB
        temp.Add(bytes[i]);     // 2nd LSB
        temp.Add(bytes[i + 1]); // 2nd MSB
        temp.Add(bytes[i + 2]); // MSB
    }

    // BitConverter requires collection to be an array.
    paddedBytes = temp.ToArray();
    temp = null;
    bytes = null;

    for (var i = 0; i < paddedBytes.Length / 4; i++)
    {
        samples.Add(BitConverter.ToInt32(paddedBytes,i * 4) / 2147483648f); // Skip the bit shift and just divide,since our sample has been "shited" 8 places to the right we need to divide by 2147483648,not 8388608.
    }

    return samples;
}

对于quick1实现,您可以执行以下操作,

private List<float> Read24BitSamples(FileStream stream,int endIndex)
{
    var bytes = ReadChannelBytes(stream,endIndex);
    var samples = new float[bytes.Length / 3];

    for (var i = 0; i < bytes.Length; i += 3)
    {
        samples[i / 3] = (bytes[i] << 8 | bytes[i + 1] << 16 | bytes[i + 2] << 24) / 2147483648f;
    }

    return samples.ToList();
}

1将上述代码与之前的方法进行基准测试后,此解决方案的速度提高了约450％至550％.