问题描述
我正在尝试正确读取 WAVE 文件、PCM、单声道、16 位(每个样本 2 个字节)。我设法阅读了标题。问题是读取(写入)数据部分。
据我所知,数据块中的 16 位样本是小端的,并且“拆分”为两个 8 位的块。所以对我来说,读取正确数据的方法应该是:
问题是我不知道如何处理小字节序以及这些样本没有无符号的事实,所以我不能使用
这是我做过的测试之一,没有成功:
int8_t buffer[],firstbyte,secondbyte;
int16_t result;
std::vector<int16_t> data;
while(Read bytes and put them in buffer){
for (int j=0;j<bytesReadFromTheFile;j+=2){
firstbyte = buffer[j];
secondbyte = buffer[j+1];
result = (firstbyte);
result = (result << 8)+secondbyte; //shift first byte and add second
data.push_back(result);
}
}
为了更详细,我使用了网上找到的这段代码并从它开始创建了一个类(过程是一样的,但是类配置很长,并且有很多功能不是' t 对这个问题很有用):
#include <iostream>
#include <string>
#include <fstream>
#include <cstdint>
using std::cin;
using std::cout;
using std::endl;
using std::fstream;
using std::string;
typedef struct WAV_HEADER
{
/* RIFF Chunk Descriptor */
uint8_t RIFF[4]; // RIFF Header Magic header
uint32_t ChunkSize; // RIFF Chunk Size
uint8_t WAVE[4]; // WAVE Header
/* "fmt" sub-chunk */
uint8_t fmt[4]; // FMT header
uint32_t Subchunk1Size; // Size of the fmt chunk
uint16_t AudioFormat; // Audio format 1=PCM,6=mulaw,7=alaw,257=IBM Mu-Law,258=IBM A-Law,259=ADPCM
uint16_t NumOfChan; // Number of channels 1=Mono 2=Sterio
uint32_t SamplesPerSec; // Sampling Frequency in Hz
uint32_t bytesPerSec; // bytes per second
uint16_t blockAlign; // 2=16-bit mono,4=16-bit stereo
uint16_t bitsPerSample; // Number of bits per sample
/* "data" sub-chunk */
uint8_t Subchunk2ID[4]; // "data" string
uint32_t Subchunk2Size; // Sampled data length
} wav_hdr;
// Function prototypes
int getFileSize(FILE* inFile);
int main(int argc,char* argv[])
{
wav_hdr wavHeader;
int headerSize = sizeof(wav_hdr),filelength = 0;
const char* filePath;
string input;
if (argc <= 1)
{
cout << "Input wave file name: ";
cin >> input;
cin.get();
filePath = input.c_str();
}
else
{
filePath = argv[1];
cout << "Input wave file name: " << filePath << endl;
}
FILE* wavFile = fopen(filePath,"r");
if (wavFile == nullptr)
{
fprintf(stderr,"Unable to open wave file: %s\n",filePath);
return 1;
}
//Read the header
size_t bytesRead = fread(&wavHeader,1,headerSize,wavFile);
cout << "Header Read " << bytesRead << " bytes." << endl;
if (bytesRead > 0)
{
//Read the data
uint16_t bytesPerSample = wavHeader.bitsPerSample / 8; //Number of bytes per sample
uint64_t numSamples = wavHeader.ChunkSize / bytesPerSample; //How many samples are in the wav file?
static const uint16_t BUFFER_SIZE = 4096;
int8_t* buffer = new int8_t[BUFFER_SIZE];
while ((bytesRead = fread(buffer,sizeof buffer[0],BUFFER_SIZE / (sizeof buffer[0]),wavFile)) > 0)
{
* /** DO SOMETHING WITH THE WAVE DATA HERE **/ *
cout << "Read " << bytesRead << " bytes." << endl;
}
delete [] buffer;
buffer = nullptr;
filelength = getFileSize(wavFile);
cout << "File is :" << filelength << " bytes." << endl;
cout << "RIFF header :" << wavHeader.RIFF[0] << wavHeader.RIFF[1] << wavHeader.RIFF[2] << wavHeader.RIFF[3] << endl;
cout << "WAVE header :" << wavHeader.WAVE[0] << wavHeader.WAVE[1] << wavHeader.WAVE[2] << wavHeader.WAVE[3] << endl;
cout << "FMT :" << wavHeader.fmt[0] << wavHeader.fmt[1] << wavHeader.fmt[2] << wavHeader.fmt[3] << endl;
cout << "Data size :" << wavHeader.ChunkSize << endl;
// display the sampling Rate from the header
cout << "Sampling Rate :" << wavHeader.SamplesPerSec << endl;
cout << "Number of bits used :" << wavHeader.bitsPerSample << endl;
cout << "Number of channels :" << wavHeader.NumOfChan << endl;
cout << "Number of bytes per second :" << wavHeader.bytesPerSec << endl;
cout << "Data length :" << wavHeader.Subchunk2Size << endl;
cout << "Audio Format :" << wavHeader.AudioFormat << endl;
// Audio format 1=PCM,259=ADPCM
cout << "Block align :" << wavHeader.blockAlign << endl;
cout << "Data string :" << wavHeader.Subchunk2ID[0] << wavHeader.Subchunk2ID[1] << wavHeader.Subchunk2ID[2] << wavHeader.Subchunk2ID[3] << endl;
}
fclose(wavFile);
return 0;
}
// find the file size
int getFileSize(FILE* inFile)
{
int fileSize = 0;
fseek(inFile,SEEK_END);
fileSize = ftell(inFile);
fseek(inFile,SEEK_SET);
return fileSize;
}
问题出在 /** DO SOMETHING WITH THE WAVE DATA HERE **/ 。我不知道如何获得样本值。
解决方法
我是 Java 程序员,不是 C++,但我经常处理这个问题。
PCM 数据按帧组织。如果它是单声道、小端、16 位,第一个字节将是值的下半部分,第二个字节将是上半部分并包括符号位。 Big-endian 将反转字节。如果是立体声,则在进入下一帧之前会完整地呈现完整帧(我认为它是左后右,但我不确定)。
我对显示的所有代码感到惊讶。在 Java 中,以下内容对于编码为有符号值的 PCM 就足够了:
public short[] fromBufferToPCM(short[] audioPCM,byte[] buffer)
{
for (int i = 0,n = buffer.length; i < n; i += 2)
{
audioPCM[i] = (buffer[i] & 0xff) | (buffer[i + 1] << 8);
}
return audioBytes;
}
IDK 如何将其直接转换为 C++,但我们只是将两个字节进行 OR 运算,第二个字节首先向左移动 8 位。纯移位拾取符号位。 (我不记得为什么包含 & 0xff 了——我很久以前写过这个并且它有效。)
好奇为什么评论中有这么多答案而不是作为答案发布。我认为评论是为了澄清 OP 的问题。