使用delphi读取.wav文件

Question

我现在尝试使用delphi读取.wav文件,这里是我的代码:

type
  TWaveHeader = packed record
    Marker_RIFF: array [0..3] of char;
    ChunkSize: cardinal;

    Marker_WAVE: array [0..3] of char;
    Marker_fmt: array [0..3] of char;
    SubChunkSize: cardinal;

    FormatTag: word;

    NumChannels: word;
    SampleRate: longint;
    BytesPerSecond: longint;
    BytesPerSample: word;
    BitsPerSample: word;


    Marker_data: array [0..3] of char;
    DataBytes: longint;
  end;

  TChannel = record
  Data : array of double;
end;

一些私人移民

private
    wavehdr:TWaveHeader;
    wavedata:array[0..3]of TChannel;
    numsamples:integer;

功能

  FillChar(wavehdr, sizeof(wavehdr), 0);
  Stream.Read(wavehdr, sizeof(wavehdr));

  { Log Header data }
  with memo1.Lines do begin
    Add('Filename : '+od.FileName);
    Add('Header size : '+inttostr(sizeof(wavehdr)));
    tmpstr := wavehdr.Marker_RIFF;
    Add('RIFF ID : '+tmpstr+'');
    Add('Chunk size : '+inttostr(wavehdr.ChunkSize));
    tmpstr := wavehdr.Marker_WAVE;
    Add('WAVE ID : '+tmpstr+'');
    tmpstr := wavehdr.Marker_fmt;
    Add('''fmt '' ID : '+tmpstr+''' ');
    Add('SubChunk size : '+inttostr(wavehdr.SubChunkSize));
    Add('Format : '+inttostr(wavehdr.FormatTag));
    Add('Num Channels : '+inttostr(wavehdr.NumChannels));
    Add('Sample rate : '+inttostr(wavehdr.SampleRate));
    Add('Bytes per second : '+inttostr(wavehdr.BytesPerSecond));
    Add('Bits per sample : '+inttostr(wavehdr.BitsPerSample));
    Add('Block Align : '+inttostr((wavehdr.NumChannels*wavehdr.BitsPerSample)div 8));
  end;

  numsamples := (file.size div (wavehdr.NumChannels*wavehdr.BitsPerSample)div 8) div wavehdr.BytesPerSample;
  case wavehdr.NumChannels of
      1:begin
        SetLength(wavedata[0].Data, numsamples);
        Stream.Read(wavedata[0].Data[0], numsamples);
      end;

      2:begin
        SetLength(wavedata[0].Data, numsamples);
        SetLength(wavedata[1].Data, numsamples);
        for i := 0 to high(wavedata[0].Data) do begin
          Stream.Read(wavedata[0].Data[i], 2);
          Stream.Read(wavedata[1].Data[i], 2);
        end;
      end;
  end;

上面的代码给出了与.wav标题完全相同的信息和细节(与MATLAB DOES相同),它是:

• 文件名:E:\ dephi\classic3.wav
• RIFF ID:RIFF
• 块大小:18312354
• WAVE ID:WAVE
• 'fmt'ID:fmt'
• SubChunk大小:16
• 格式:1(PCM)
• Num频道:2(立体声)
• 采样率:44100
• 每秒字节数:176400
• 每个样本的比特:16
• Block Align:4

除了我计算的总样本数据(波数据的大小/波数据的块对齐)-44,44是wav的标题.它不准确,有时会错过5,1,10.我只测试了5个样本.这里有一个例子:

• classic1.wav matlab:3420288,delphi(我的计算):( 13681352/4)-44 = 3420294
• classic2.wav matlab:2912256,delphi(我的计算):( 11649204/4)-44 = 2912257

而且来自matlab和delphi的样本数据值也不同

classic1.wav MATLAB :(前10个值为leftchannel和rightchannel)

1. -3.05175781250000e-05 [] 6.10351562500000e-05
2. -6.10351562500000e-05 [] 6.10351562500000e-05
3. -6.10351562500000e-05 [] 3.05175781250000e-05
4. 0 [] -3.05175781250000e-05
5. 6.10351562500000e-05 [] -6.10351562500000e-05
6. 6.10351562500000e-05 [] -6.10351562500000e-05
7. 3.05175781250000e-05 [] -3.05175781250000e-05
8. 6.10351562500000e-05 [] -6.10351562500000e-05
9. 3.05175781250000e-05 [] 0
10. -3.05175781250000e-05 [] 6.10351562500000e-05

DELPHI :(前10个值为leftchannel和rightchannel)

1. 9.90156960830442E-320 [] 1.00265682167023E-319
2. 9.90156960830442E-320 [] 9.77113627780233E-320
3. 3.26083326255223E-322 [] 0
4. 1.39677298735779E-319 [] 1.37088394751571E-319
5. 1.45932169812129E-319 [] 1.33373021094845E-319
6. 1.23175506164681E-319 [] 1.206903559661E-319
7. 1.28239679034554E-319 [] 1.40932225476216E-319
8. 1.37068632125737E-319 [] 1.33382902407761E-319
9. 1.33373021094845E-319 [] 1.25685359645555E-319
10. 1.40907522193924E-319 [] 1.33358199125469E-319

我的问题是:

1. 查找wav文件的总样本时,如何正确执行？
2. matlab和delphi以不同的方式读取wav文件(数据块)的方式是什么？或者我的代码可能是错误的？
3. 有没有办法像MATLAB一样获得相同的值？

编辑:我跟着mBo建议并将其改为mbo建议

Data : array of SmallInt;
numsamples := wavehdr.DataBytes div (wavehdr.NumChannels * wavehdr.BitsPerSample div 8);
Stream.Read(wavedata[0].Data[i], SizeOf(SmallInt));

口译部分我不确定,但我把它改成了

floattostr(wavedata[0].Data[i]/32768.0)
floattostr(wavedata[1].Data[i]/32768.0)

结果我得到:

1. 0.611602783203125 [] 0.61932373046875
2. 0.611602783203125 [] 0.603546142578125
3. 0.0023193359375 [] 0
4. 0.862762451171875 [] 0.846771240234375
5. 0.901397705078125 [] 0.823822021484375
6. 0.760833740234375 [] 0.7454833984375
7. 0.7921142578125 [] 0.870513916015625
8. 0.799774169921875 [] 0.761016845703125
9. 0.8238525390625 [] 0.782623291015625
10. 0.354766845703125 [] 0.76123046875

Answer 1

Wav文件(每个样本的位数:16)包含带符号的16位整数数据(SmallInt类型),但是您以float 8字节类型的Double数组读取数据.

你可以申报

Data : array of SmallInt;

计算

numsamples := wavehdr.DataBytes div (wavehdr.NumChannels * wavehdr.BitsPerSample div 8);

读它们为

Stream.Read(wavedata[0].Data[0], numsamples * SizeOf(SmallInt))
or multichannel case:
Stream.Read(wavedata[0].Data[i], SizeOf(SmallInt));

然后将数据值解释为浮点数Data [i]/32768.0

注意,matlab值3.05175781250000e-05 = 1/32768.0是16位信号的最小量子