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March 3, 2018 22:22
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init | |
while true | |
if audout does not contain buffer1 | |
decode to buffer1, then schedule it | |
if audout does not contain buffer2 | |
decode to buffer2, then schedule it. | |
wait for buffer finish. make sure timeout is a reasonable time for 3 buffers worth of audio to play in, just in case. | |
-- Modified from Ghabry's code | |
instance->LockMutex(); | |
for (int index = 0; index < 2; index++) { | |
// If the function errors, we assume the buffer isn't there. Might want to change this. | |
bool contain = true; | |
// Either one or none of the buffers are playing, in which case this appends the buffers that aren't playing, | |
// or two buffers are playing due to some sort of freak accident, | |
audoutContainsAudioOutBuffer(source_buffer[index], &contain); | |
if (!contain) { | |
instance->Decode((uint8_t*)source_buffer[index]->buffer, buf_size); | |
audoutAppendAudioOutBuffer(source_buffer[index]); | |
} | |
} | |
instance->UnlockMutex(); | |
// we don't do anything with the result values, but they have to be non-null | |
// Apparently, the timeout is documented as waiting until all buffers are finished. | |
// If this is true, comment this out, add a sleep for 1/2 buffer time | |
audoutWaitPlayFinish(&released_buffer, &released_count, U64_MAX); |
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