audacia
/
audacia
1
0
Fork 0
Code Issues Pull Requests Projects Releases Wiki Activity

As mentioned in commit message for 12574, here is one example of a routine 

that was doing extra processing because Dither couldn't return interleaved
buffers.
This commit is contained in:
lllucius 2013-09-22 06:26:06 +00:00
parent a3d27b135b
commit 7ddd4ed2ab
1 changed files with 10 additions and 16 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

26
src/AudioIO.cpp
View File

@ -3217,25 +3217,19 @@ static void DoSoftwarePlaythrough(const void *inputBuffer,
float *outputBuffer,
int len)
{
float *tempBuffer = (float *)alloca(len * sizeof(float));
int i, j;
for (int i=0; i < inputChannels; i++) {
samplePtr inputPtr = ((samplePtr)inputBuffer) + (i * SAMPLE_SIZE(inputFormat));
samplePtr outputPtr = ((samplePtr)outputBuffer) + (i * SAMPLE_SIZE(floatSample));
for(j=0; j<inputChannels; j++) {
samplePtr inputPtr = ((samplePtr)inputBuffer) + (j * SAMPLE_SIZE(inputFormat));
CopySamples(inputPtr, inputFormat,
(samplePtr)tempBuffer, floatSample,
len, true, inputChannels);
for(i=0; i<len; i++)
outputBuffer[2*i + (j%2)] = tempBuffer[i];
// One mono input channel goes to both output channels...
if (inputChannels == 1)
for(i=0; i<len; i++)
outputBuffer[2*i + 1] = tempBuffer[i];
(samplePtr)outputPtr, floatSample,
len, true, inputChannels, 2);
}
// One mono input channel goes to both output channels...
if (inputChannels == 1)
for (int i=0; i < len; i++)
outputBuffer[2*i + 1] = outputBuffer[2*i];
}
int audacityAudioCallback(const void *inputBuffer, void *outputBuffer,