Re: Dealing with MONO12 images in memory

On Tue, 21 Nov 2017, Robert wrote:

> Not sure about the proper use of the 3rd argument in the image.read() function - I've tried both 16 and 2.

For what you are trying, the third argument should be 16.  There is a 
four argument variant:

Read single image frame of specified size, depth, and format from 
in-memory Blob:

void            read ( const Blob         &blob_,
                        const Geometry     &size_,
                        const unsigned int depth_,
                        const std::string  &magick_ )

which accepts the format specifier (which you need!).

For example

   image.read(blob,Geometry(arv_width, arv_height), 16, "GRAY");

Note that 16-bit values could be big-endian or little-endian.  The 
default is big-endian but little-endian is what Intel CPUs use.

Another way to read it would then be:

   image.depth(16);
   image.endian(LSBEndian);
   image.read(blob,Geometry(arv_width, arv_height));

Lastly, don't pass up this form of read:

Read single image frame from an array of raw pixels, with specified 
storage type (ConstituteImage), e.g. image.read( 640, 480, "RGB", 0, 
pixels ):

void            read ( const unsigned int width_,
                        const unsigned int height_,
                        const std::string &map_,
                        const StorageType  type_,
                        const void        *pixels_ )

This is a (sometimes) less CPU-efficient but more free-form way to 
read pixels from raw 16-bit data expressed in the host's native 
format:

image.read(arv_width, arv_height, "I", ShortPixel, pRaw16Image);

The underlying C function (with full documentation) is documented at 
"http://www.graphicsmagick.org/api/constitute.html#dispatchimage".  It 
seems that the C function offers a bit more power than the C++ 
equivalent.

Bob
-- 
Bob Friesenhahn
[email protected], http://www.simplesystems.org/users/bfriesen/
GraphicsMagick Maintainer,    http://www.GraphicsMagick.org/

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