Most efficient way to determine the number of non-zero entries of a CImg?

[nji9nji9]

Is there a member function to use, or do I have to iterate
pixel-by-pixel myself?
Probably code it with vector ops at the internal structure...?

[dtschump]

I'd say :

  const unsigned int nb_non_zeros = img.size() - img.get_histogram(1,0,0)[0];

A bit tricky, but concise enough :)