用BinaryReader解析Wave文件

Question

在.NET Assembly mscorlib System.IO命名空间中，我使用ReadInt16()方法遍历音频数据字节，并将有符号整数值转储到一个文本文件中。如何解释与一个采样率相关的两个值？也就是说，如果我有1秒的单声道数据，那么将有88200个字节，因此使用ReadInt16()将返回88200个离散整数。这是太多的信息，我应该只有44100个整数。那么，我是否需要使用不同的方法，或者可能在每次迭代中将循环前进1。

许多thanks..........Mickey

Answer 1

using System;
using System.IO;

public struct WaveFormat
{

    private short m_FormatTag;       // most often PCM = 1      
    private short m_nChannels;       // number of channels      
    private int m_SamplesPerSecond;  // samples per second eg 44100     
    private int m_AvgBytesPerSecond; // bytes per second eg 176000      
    private short m_BlockAlign;      // blockalign (byte per sample) eg 4 bytes         
    private short m_BitsPerSample;   // bits per sample, 8, 16, 24

    public WaveFormat(byte BPS, int SPS, byte nChn)
    {
        m_FormatTag = 1; //PCM
        m_nChannels = nChn;
        m_SamplesPerSecond = SPS;
        m_BitsPerSample = BPS;
        m_BlockAlign = (short)(m_nChannels * m_BitsPerSample / 8);
        m_AvgBytesPerSecond = (int)(m_BlockAlign * m_SamplesPerSecond);
    }
    public short FormatTag
    {
        get { return m_FormatTag; }
        set { m_FormatTag = value; }
    }
    public short Channels
    {
        get { return m_nChannels; }
    }
    public int SamplesPerSecond
    {
        get { return m_SamplesPerSecond; }
    }
    public int AvgBytesPerSecond
    {
        get { return m_AvgBytesPerSecond; }
    }
    public short BlockAlign
    {
        get { return m_BlockAlign; }
    }
    public short BitsPerSample
    {
        get { return m_BitsPerSample; }
    }
    public void Read(BinaryReader br)
    {
        m_FormatTag = br.ReadInt16();
        m_nChannels = br.ReadInt16();
        m_SamplesPerSecond = br.ReadInt32();
        m_AvgBytesPerSecond = br.ReadInt32();
        m_BlockAlign = br.ReadInt16();
        m_BitsPerSample = br.ReadInt16();
    }
    public void Write(BinaryWriter bw)
    {
        bw.Write(m_FormatTag);
        bw.Write(m_nChannels);
        bw.Write(m_SamplesPerSecond);
        bw.Write(m_AvgBytesPerSecond);
        bw.Write(m_BlockAlign);
        bw.Write(m_BitsPerSample);
    }
    public override string ToString()
    {
        System.Text.StringBuilder sb = new System.Text.StringBuilder();
        sb.AppendLine("FormatTag:         " + m_FormatTag.ToString());
        sb.AppendLine("nChannels:         " + m_nChannels.ToString());
        sb.AppendLine("SamplesPerSecond:  " + m_SamplesPerSecond.ToString());
        sb.AppendLine("AvgBytesPerSecond: " + m_AvgBytesPerSecond.ToString());
        sb.AppendLine("BlockAlign:        " + m_BlockAlign.ToString());
        sb.AppendLine("BitsPerSample:     " + m_BitsPerSample.ToString());
        return sb.ToString();
    }
}

Answer 2

通常，当您读取数据数组时，代码应如下所示：

for(int i = 0; i < totalNumberOfEntries; i++)
{
  // read all data for this entry
  var component1 = reader.ReadXXX();
  var component2 = reader.ReadXXX();

  // deal with data for this entry
  someEntryStroage.Add(new Entry(component1, component2);
}

在你的例子中，最有可能的情况是(我不知道Int16文件格式)，你要么需要读取成对的Wave值(如果样本在一起)，要么需要单独读取通道，如果一个通道的数据是另一个通道的数据。

Answer 3

你一定要读一下chunkinfos。data-chunk告诉你需要读取多少字节。WaveFormat会告诉你每秒钟的平均字节数，甚至更多。我有一些VB代码。