Question

Let’s say I have an image represented by a CFMutableBitVector. That means that ever 0 bit represents an off pixel, and every 1 bit represents an on pixel.

Is there a way to, easily and quickly convert this CFMutableBitVector to a low level array of ints where an on-bit represents an all-on-bit integer, and an off-bit represents an all-off-bit integer?

The CFMutableBitVector [0,0,0,1,1,0,1,0];

would translate to:

int[8] =
[
    00000000000000000000000000000000,
    00000000000000000000000000000000,
    00000000000000000000000000000000,
    11111111111111111111111111111111,
    11111111111111111111111111111111,
    00000000000000000000000000000000,
    11111111111111111111111111111111,
    00000000000000000000000000000000,
];

Now, this being an image from a video stream, this will have to happen very quickly — What’s the fastest way to do such a conversion in Objective-C?

Answer 1

Here’s the sample conversion …

  u_int32_t sample = UINT32_MAX;
  CFBitVectorRef bv = CFBitVectorCreate( NULL, ( const u_int8_t * )&sample, sizeof( u_int32_t ) * 8 );

  CFIndex bitCount = CFBitVectorGetCount( bv );
  u_int32_t *a = malloc( bitCount * sizeof( u_int32_t ) );
  for ( CFIndex i = 0 ; i < bitCount ; i++ ) {
    a[i] = CFBitVectorGetBitAtIndex( bv, i ) * UINT32_MAX;
  }

  for ( int i = 0 ; i < bitCount ; i++ ) {
    NSLog( @"Index: %d Number: %u", i, a[i] );
  }
  free( a );