问题描述
我正在使用双簧管在Android上播放声音文件。我想同时播放44.1kHz和48kHz的文件,因此需要重新采样。
解码和播放文件效果很好,但是由于我有两个不同的采样率,因此我需要重新采样(由于音频流为48kHz,我目前正在尝试44.1至48。)
因此,我尝试使用oboe's resampler进行重采样,但是我无法完全理解该怎么做。按照自述文件指南转换固定数量的输入帧(我认为这是我必须做的吗?),我尝试实现如下。代码的第一部分获取解码并在采样率相等时返回(这部分按预期工作),第二部分是在必要时我尝试重新采样的地方:
StorageDataSource *StorageDataSource::newFromStorageAsset(AMediaExtractor &extractor,const char *fileName,AudioProperties targetProperties) {
std::ifstream stream;
stream.open(fileName,std::ifstream::in | std::ifstream::binary);
stream.seekg(0,std::ios::end);
long size = stream.tellg();
stream.close();
constexpr int kMaxCompressionRatio{12};
const long maximumDataSizeInBytes =
kMaxCompressionRatio * (size) * sizeof(int16_t);
auto decodedData = new uint8_t[maximumDataSizeInBytes];
int32_t rate = NDKExtractor::getSampleRate(extractor);
int32_t *inputSampleRate = &rate;
int64_t bytesDecoded = NDKExtractor::decode(extractor,decodedData,targetProperties);
auto numSamples = bytesDecoded / sizeof(int16_t);
auto outputBuffer = std::make_unique<float[]>(numSamples);
// The NDK decoder can only decode to int16,we need to convert to floats
oboe::convertPcm16ToFloat(
reinterpret_cast<int16_t *>(decodedData),outputBuffer.get(),bytesDecoded / sizeof(int16_t));
if (*inputSampleRate == targetProperties.sampleRate) {
return new StorageDataSource(std::move(outputBuffer),numSamples,targetProperties);
} else {
// this is where I try to convert the sample rate
float *inputBuffer;
inputBuffer = reinterpret_cast<float *>(decodedData); // is this correct?
float *outputBuffer2; // multi-channel buffer to be filled,Todo improve name
int numInputFrames; // number of frames of input
// Todo is this correct?
numInputFrames = numSamples / 2;
int numOutputFrames = 0;
int channelCount = 2;
resampler::MultiChannelResampler *mResampler = resampler::MultiChannelResampler::make(
2,// channel count
44100,// input sampleRate
48000,// output sampleRate
resampler::MultiChannelResampler::Quality::Best); // conversion quality
int inputFramesLeft = numInputFrames;
while (inputFramesLeft > 0) {
if (mResampler->isWriteNeeded()) {
mResampler->writeNextFrame(inputBuffer);
inputBuffer += channelCount;
inputFramesLeft--;
} else {
mResampler->readNextFrame(outputBuffer2);
outputBuffer2 += channelCount;
numOutputFrames++;
}
}
delete mResampler;
// return is missing!
}
// returning the original data since above code doesn't work properly yet
return new StorageDataSource(std::move(outputBuffer),targetProperties);
}
重新采样会因SIGSEV
而崩溃:
A: signal 11 (SIGSEGV),code 1 (SEGV_MAPERR),fault addr 0x7fe69c7000
A: x0 0000007c0e3d1e00 x1 0000007fe69c7000 x2 0000007bb77dd198 x3 0000007bf5432140
A: x4 0000000000000021 x5 8080800000000000 x6 fefeff7b976e0667 x7 7f7f7f7fff7f7f7f
A: x8 0000000000000660 x9 0000000000000660 x10 0000000000000000 x11 0000007bf5435840
A: x12 0000007bb77dd118 x13 0000000000000008 x14 0000007bf54321c0 x15 0000000000000008
A: x16 0000007bf5432200 x17 0000000000000000 x18 0000007fe69bf7ba x19 0000007c14e14c00
A: x20 0000000000000000 x21 0000007c14e14c00 x22 0000007fe69c0d70 x23 0000007bfc6e5dc7
A: x24 0000000000000008 x25 0000007c9b7705f8 x26 0000007c14e14ca0 x27 0000000000000002
A: x28 0000007fe69c0aa0 x29 0000007fe69c0420
A: sp 0000007fe69c0400 lr 0000007bf94f61f0 pc 0000007bf9501b5c
A: backtrace:
A: #00 pc 0000000000078b5c /data/app/myapp-G-GmPWmPgOGfffk-qHsQxw==/lib/arm64/libnative-lib.so (resampler::polyphaseResamplerStereo::readFrame(float*)+684)
A: #01 pc 000000000006d1ec /data/app/myapp-G-GmPWmPgOGfffk-qHsQxw==/lib/arm64/libnative-lib.so (resampler::MultiChannelResampler::readNextFrame(float*)+44)
A: #02 pc 000000000006c84c /data/app/myapp-G-GmPWmPgOGfffk-qHsQxw==/lib/arm64/libnative-lib.so (StorageDataSource::newFromStorageAsset(AMediaExtractor&,char const*,AudioProperties)+1316)
A: #03 pc 78bbcdd7f9b20dbe <unkNown>
这是我的主要问题: 首先,如何正确获取输入的帧数?帧如何正确处理音频数据?我确实对此进行了研究,但仍不确定是否能得到这个?这是一个常数吗?如何计算帧数。它与样本,采样率和比特率有什么关系?
第二,我是否完全使用了正确的输入数据?我使用我的decodedData
值,因为那是我从解码器取回的值,只是将其reinterpret_cast
返回到float*
。
由于我对C ++相当缺乏经验,所以我不确定自己所做的是否正确,并且我可能会在这段代码中引入多个错误。
编辑:由于我正在尝试对解码后的输出进行重新采样,因此我假设来自here的有关PCM的这一信息在这里解释了帧的含义:
For encodings like PCM,a frame consists of the set of samples for all channels at a given point in time,and so the size of a frame (in bytes) is always equal to the size of a sample (in bytes) times the number of channels.
我的情况是否正确?那意味着我可以从采样数,音频位的长度和通道数中减去帧数吗?
解决方法
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