RtAudio: understanding frame count vs buffer size

Question

Looking at the duplex example from the documentation of RtAudio, I do not understand why the number of frames has to be multiplied by 8 to get the good buffer size to copy.

In the example there is bufferBytes = bufferFrames * 2 * 4;. I imagine the * 2 refers to the number of channels, but I see no reason for the * 4.

Whole example:

#include "RtAudio.h"
#include <iostream>
#include <cstdlib>
#include <cstring>
// Pass-through function.
int inout( void *outputBuffer, void *inputBuffer, unsigned int nBufferFrames,
           double streamTime, RtAudioStreamStatus status, void *data )
{
  // Since the number of input and output channels is equal, we can do
  // a simple buffer copy operation here.
  if ( status ) std::cout << "Stream over/underflow detected." << std::endl;
  unsigned int *bytes = (unsigned int *) data;
  memcpy( outputBuffer, inputBuffer, *bytes );
  return 0;
}
int main()
{
 RtAudio adac;
  if ( adac.getDeviceCount() < 1 ) {
    std::cout << "\nNo audio devices found!\n";
    exit( 0 );
  }
  // Set the same number of channels for both input and output.
  unsigned int bufferBytes, bufferFrames = 512;
  RtAudio::StreamParameters iParams, oParams;
  iParams.deviceId = 0; // first available device
  iParams.nChannels = 2;
  oParams.deviceId = 0; // first available device
  oParams.nChannels = 2;
  try {
    adac.openStream( &oParams, &iParams, RTAUDIO_SINT32, 44100, &bufferFrames, &inout, (void *)&bufferBytes );
  }
  catch ( RtAudioError& e ) {
    e.printMessage();
    exit( 0 );
  }

  // THIS HERE...
  //
  bufferBytes = bufferFrames * 2 * 4; // <---- WHY * 4??
  //
  //

  try {
    adac.startStream();
    char input;
    std::cout << "\nRunning ... press <enter> to quit.\n";
    std::cin.get(input);
    // Stop the stream.
    adac.stopStream();
  }
  catch ( RtAudioError& e ) {
    e.printMessage();
    goto cleanup;
  }
 cleanup:
  if ( adac.isStreamOpen() ) adac.closeStream();
  return 0;
}