我正在将FLAC音频解码到内存中,并将解码的音频数据传递到OpenAL:void alBufferData (ALuint bufferName, ALenum format, const ALvoid *data, ALsizei size, ALsizei frequency);
来自解码音频的数据进入矿井std::vector<FLAC__int32> data_;
。我正试图将LEFT(左)和RIGHT(右)通道封装到其中(AL_FORMAT_STERIO16)。但是,我不明白如何在data_
矢量中存储/对齐这些通道。
所以我有了libFLAC虚拟回调成员函数:
FLAC__StreamDecoderWriteStatus
Source::write_callback (
FLAC__Frame const* _frame, FLAC__int32 const *const _buffer[])
{
for(size_t i(0); i < _frame->header.blocksize; i++) {
data_[index_] = _buffer[0][i]; // channel audio on the left
++index_;
data_[index_] = _buffer[1][i]; // what about the right channel?
} // jump
return FLAC__STREAM_DECODER_WRITE_STATUS_CONTINUE;
} // main
目前,在音频播放过程中,我只听到LEFT频道。在声音播放完毕后会有静态声音,我认为这是丢失的右声道数据。我如何让正确的频道也工作?
此外,这是根据libFLAC:的元数据回调签名
void
Source::metadata_callback (const ::FLAC__StreamMetadata *metadata)
{
total_samples_ = metadata->data.stream_info.total_samples;
rate_ = metadata->data.stream_info.sample_rate;
channels_ = metadata->data.stream_info.channels;
bps_ = metadata->data.stream_info.bits_per_sample;
switch (bps_) {
case 16 :
if (channels_ > 1) {
format_ = AL_FORMAT_STEREO16; } else {
format_ = AL_FORMAT_MONO16; }
break;
case 8 :
if (channels_ > 1) {
format_ = AL_FORMAT_STEREO8; } else {
format_ = AL_FORMAT_MONO8; }
break;
default:
break;
}
size_ = (ALuint)(rate_ * channels_ * (bps_ / 8));
data_.resize(total_samples_); index_ = 0;
} // main
一个有效的解决方案是将以下struct
分析为矢量数据类型,如下所示:
struct Data
{
FLAC__int16 channelLeft_;
FLAC__int16 channelRight_;
};
std::vector<Source::Data> data_;
而不是像这样分配size_
:
size_ = total_samples_ * sizeof(Source::Data);
最后,数据循环现在应该是:
for(size_t i(0); i < _frame->header.blocksize; i++) {
data_[index_].channelLeft_ = _buffer[0][i];
data_[index_].channelRight_ = _buffer[1][i];
++index_;
} // jump