c - 将 2 通道 WAV 文件解交织成两个包含原始数据的文本文件

Question

我有一个基于上述问题的任务。每个样本的采样频率和大小在问题中是已知的。我只需要了解为此所需的编码类型。

Answer 1

使用文件格式规范（例如这个）来查看如何读取文件头、确定采样率、比特率等。

The canonical WAVE format starts with the RIFF header:

0         4   ChunkID          Contains the letters "RIFF" in ASCII form
                               (0x52494646 big-endian form).
4         4   ChunkSize        36 + SubChunk2Size, or more precisely:
                               4 + (8 + SubChunk1Size) + (8 + SubChunk2Size)
                               This is the size of the rest of the chunk 
                               following this number.  This is the size of the 
                               entire file in bytes minus 8 bytes for the
                               two fields not included in this count:
                               ChunkID and ChunkSize.
8         4   Format           Contains the letters "WAVE"
                               (0x57415645 big-endian form).

The "WAVE" format consists of two subchunks: "fmt " and "data":
The "fmt " subchunk describes the sound data's format:

12        4   Subchunk1ID      Contains the letters "fmt "
                               (0x666d7420 big-endian form).
16        4   Subchunk1Size    16 for PCM.  This is the size of the
                               rest of the Subchunk which follows this number.
20        2   AudioFormat      PCM = 1 (i.e. Linear quantization)
                               Values other than 1 indicate some 
                               form of compression.
22        2   NumChannels      Mono = 1, Stereo = 2, etc.
24        4   SampleRate       8000, 44100, etc.
28        4   ByteRate         == SampleRate * NumChannels * BitsPerSample/8
32        2   BlockAlign       == NumChannels * BitsPerSample/8
                               The number of bytes for one sample including
                               all channels. I wonder what happens when
                               this number isn't an integer?
34        2   BitsPerSample    8 bits = 8, 16 bits = 16, etc.
          2   ExtraParamSize   if PCM, then doesn't exist
          X   ExtraParams      space for extra parameters

The "data" subchunk contains the size of the data and the actual sound:

36        4   Subchunk2ID      Contains the letters "data"
                               (0x64617461 big-endian form).
40        4   Subchunk2Size    == NumSamples * NumChannels * BitsPerSample/8
                               This is the number of bytes in the data.
                               You can also think of this as the size
                               of the read of the subchunk following this 
                               number.
44        *   Data             The actual sound data.

在那之后，你会发现原始 pcm 数据，像交错一样

[sample 1      ][sample 2      ]
[s1,ch1][s1,ch2][s2,ch1][s2,ch2]

您可以以写入二进制模式打开每个样本的文本文件，然后循环遍历音频数据，读取单个样本/通道的字节，然后使用fprintf或fwrite将它们写入正确的文件。

Answer 2

采样频率与此无关，但每个样本的大小（通常每个样本每个通道 8 或 16 位）决定了您需要使用的指针大小，因此这里是每个通道 8 位的示例：

char* reader = begin; // interleaved
char* left = malloc(numsamples); // de-interleaved
char* right = malloc(numsamples);
while(reader<end) {
    *left = *reader;
    ++left;
    ++reader;
    *right = *reader;
    ++right;
    ++reader;
}

要对 2 通道 16 位交错音频执行相同操作，只需将所有 3 个缓冲区声明short*为malloc(numsamples*2)

Answer 3

假设您已将 WAV 数据加载到内存中，您需要做的就是：

1. 打开两个输出文件（使用fopen()）。
2. 循环遍历样本数据，并为每个样本：
   • 将左声道的值放在第一个文件中
   • 将正确通道的值放在第二个文件中
3. 关闭文件。