将2通道WAV文件解交换为包含原始数据的两个文本文件
我根据上述问题进行了分配。 在该问题中将知道每个样本的采样频率和大小。 我只需要了解这需要的编码类型。
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）决定您需要使用哪个指针大小，因此这里每个通道8位的示例：
char* reader = begin; // interleaved char* left = malloc(numsamples); // de-interleaved char* right = malloc(numsamples); while(reader 要对2通道16位交错音频执行相同操作，您只需将所有3个缓冲区声明为short* ，而不是malloc(numsamples*2)
使用这样的文件格式规范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 (ie 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.
之后，你会发现原始pcm数据，交错像
[sample 1 ][sample 2 ] [s1,ch1][s1,ch2][s2,ch1][s2,ch2]
您可以在写入，二进制模式下打开每个样本的文本文件，然后循环播放音频数据，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。
假设您已将WAV数据加载到内存中，您需要做的就是：
- 打开两个输出文件（使用
fopen()）。 - 循环遍历样本数据，并为每个样本：
- 将左通道的值放在第一个文件中
- 将右通道的值放在第二个文件中
- 关闭文件。