We first bin the channels then we calculate the difference between pairs of channels in a datastream. E.g. For binning, if we bin 4 samples in each channel we will have 4 times less samples per channel E.g. For subtracting if we have 4 channels we end up with 2 channels. The channels will be channel_1 - channel_2 channel_3 - channel_4 This is the same as combining binning and subtracting channels. This will not work if you provide single channel data! More...

#include <BaseConverter.h>

Inheritance diagram for ChannelBinDiffT< T >:

Public Member Functions
	ChannelBinDiffT ()=default

	ChannelBinDiffT (int binSize, int channels, bool average)

	~ChannelBinDiffT ()

void	clear ()

size_t	convert (uint8_t *src, size_t size)

size_t	convert (uint8_t target, uint8_t src, size_t size)

void	reset ()

void	setAverage (bool average)

void	setBinSize (int binSize)

void	setChannels (int channels)

Protected Types
using	SumT = typename AppropriateSumType< T >::type

Protected Member Functions
bool	isConfigValid ()

void	resetState ()

void	resizeState ()

Protected Attributes
bool	average = true

int	binSize = 4

int	channels = 2

SumT *	partialBin = nullptr

int	partialBinSize = 0

Detailed Description

template<typename T = int16_t>
class audio_tools::ChannelBinDiffT< T >

We first bin the channels then we calculate the difference between pairs of channels in a datastream. E.g. For binning, if we bin 4 samples in each channel we will have 4 times less samples per channel E.g. For subtracting if we have 4 channels we end up with 2 channels. The channels will be channel_1 - channel_2 channel_3 - channel_4 This is the same as combining binning and subtracting channels. This will not work if you provide single channel data!

Author: Urs Utzinger

Template Parameters

T