从wavfile标题读取采样率

Question

我正在编写读取波形文件的代码。

我使用this document作为我的指导原则。

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的采样率。

当我读取wave文件时，我得到一个采样率值-21436。我很确定我的代码读取一个小端的整数是正确的。我确定我的测试wavfile是正确的。所以，现在我不知道为什么读取的采样率是不正确的....

我的int读取代码如下。

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

读取波形文件是如下的功能...

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有十六进制值44ac（无符号INT16）和-21436也有十六进制值44ac（签署INT16） - 中问题在于编译器在移位之前将每个带符号的char隐式转换为有符号整数。您可避免通过铸造如下（其中输出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; 
} 

或简单地读成无符号字节 - 这将避免对其他领域一样的问题：

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; 
}