Question

我正在编写代码以读取波形文件。

我使用this document作为我的指南。

它指定标题的字节22是波形文件的通道数，标题的字节24是采样率。

我正在使用由Ableton输出的测试文件作为2通道16位44100hz。我已经确认测试波形文件的格式是大胆的，以确保它确实是44100hz采样率。

当我读入波形文件时，我得到-21436的采样率值。我很确定我的代码读取一个小的endian整数是正确的。我确信我的测试wavfile是正确的。所以现在我不知道为什么读取采样率不正确....

我的int阅读代码如下。

int ReadInt(char* bytes , int start) { return (bytes[start+3] << 24) + (bytes[start+2] << 16) + (bytes[start+1] << 8) + bytes[start]; }

读取波形文件的功能如下......

WavFile::WavFile(std::string filename)
{
    std::ifstream ifs;      
    ifs.open( filename, std::ios::binary | std::ios::in );

    LogStream(LOG_DEBUG) << "WavFile::WavFile - - - - BEGIN READING WAV - - - -";

    if(ifs.fail())
        throw std::invalid_argument("WavFile::WavFile : Failed to open wavFile "+filename);

    char hbytes[HEADER_SIZE];

    ifs.read(hbytes , HEADER_SIZE);

    // check that this is actually a wave file
    bool valid_riff = hbytes[0]=='R' && hbytes[1]=='I' && hbytes[2]=='F' && hbytes[3]=='F';
    bool valid_wave = hbytes[8]=='W' && hbytes[9]=='A' && hbytes[10]=='V' && hbytes[11]=='E';
    bool valid_ftm = (hbytes[12]=='f' && hbytes[13]=='m' && hbytes[14]=='t' && hbytes[15]==' ');
    bool valid_data = (hbytes[36]=='d' && hbytes[37]=='a' && hbytes[38]=='t' && hbytes[39]=='a');

    LogStream(LOG_DEBUG) << "WavFile::WavFile - valid_riff="<<valid_riff<<" valid_wave="<<valid_wave<<" valid_ftm="<<valid_ftm<<" valid_data="<<valid_data;
    if(!(valid_data && valid_ftm && valid_riff))
        throw std::invalid_argument("WavFile::WavFile : Invalid argument - unable to open wavfile "+filename);

    int audioFormat = ReadShort(hbytes , 20);
    int SubChunk1Size = ReadInt(hbytes , 16);

    if(audioFormat != 1 || SubChunk1Size != 16)
        throw std::invalid_argument("WavFile::WavFile : Only uncompressed PCM wave format supported."+filename);    

    int subChunk2size = ReadInt(hbytes , 40);
    m_header.num_channels = ReadShort(hbytes , 22);
    m_header.sample_rate = ReadInt(hbytes , 24);
    m_header.bits_per_sample = ReadShort(hbytes , 34);

    LogStream(LOG_DEBUG) << "WavFile::WavFile num_channels="<<m_header.num_channels << " sample_rate="<<m_header.sample_rate<<" bits_per_sample="<<m_header.bits_per_sample;

    m_pcm_data.resize( subChunk2size / sizeof(int16_t) );

    LogStream(LOG_DEBUG) << "WavFile::WavFile - subChunk2size = "<<subChunk2size;
    LogStream(LOG_DEBUG) << "WavFile::WavFile - m_pcm_data.size() = "<<m_pcm_data.size();
    ifs.read((char*)m_pcm_data.data() , subChunk2size);

    LogStream(LOG_DEBUG) << "WavFile: ifstream failbit="<<ifs.fail()<<" badbit="<<ifs.bad()<<" goodbit="<<ifs.good();

    ifs.close();

    LogStream(LOG_DEBUG) << "WavFile::WavFile - - - - END READING WAV - - - -\n";
    LogStream(LOG_DEBUG) << "WavFile::WavFile";
}

Answer 1

44100具有十六进制值44ac（unsigned int16），而-21436也具有十六进制值44ac（signed int16） - 问题在于编译器在移位之前将每个带符号的字符隐式转换为有符号整数的方式。您可以通过如下转换来避免这种情况（输出44100）：

int main()
{
    char bytes[4] = { 0x44, 0xac, 0x00, 0x00 };

    printf("%i\n", (((unsigned char)bytes[3]) << 24) | (((unsigned char)bytes[2]) << 16) | (((unsigned char)bytes[1]) << 8) | ((unsigned char)bytes[0]));
    return 0;
}

或者只是作为无符号字节读取 - 这将避免其他字段出现同样的问题：

int main()
{
    unsigned char bytes[4] = { 0x44, 0xac, 0x00, 0x00 };

    printf("%i\n", (bytes[3] << 24) | (bytes[2] << 16) | (bytes[1] << 8) | bytes[0]);
    return 0;
}

从wavfile头读取samplerate

1 个答案: