This maps to 2 or 3 on an 8 bit screen, and even with high gamma I couldn't see the difference. Took me a while to clue in, but since the screen only displays in 8 bits, even the maximum value of (in my case) 968 was too dark. I tried inverting the bits, but then I got a black image. However, instead of having one long gradient, I was getting several narrow gradients instead. The intent was to have a gradient that faded in from black on the left to whatever the max x was on the right. I started with the code above, using only 'x' as the color, not 'x + y'. The color = x + y code above probably didn't produce values that were bright enough. Practically speaking, values that are several multiples of 256 should be used to see anything at all. I ran into similar problems, and I thought it would be helpful to post how I resolved the issue.Ī) The bytes have to be provided to the library MSB first, so it works if you flip the lines above to this: *row++ = (png_byte)(color > 8) ī) To actually see a 16 bit value on an 8-bit screen, any values under 256 will simply be clipped to black.
16 bit gold sparkle png how to#
Sorry for resurrecting an old thread, but I got here after googling for how to write 16 bit grayscale images. png_write_image(png_ptr, row_pointers) Png_write_png(png_ptr, info_ptr, PNG_TRANSFORM_IDENTITY, NULL) Png_set_rows(png_ptr, info_ptr, row_pointers) Row_pointers = (png_bytepp)png_malloc(png_ptr, PNG_INTERLACE_NONE, PNG_COMPRESSION_TYPE_DEFAULT, PNG_FILTER_TYPE_DEFAULT) Png_destroy_write_struct(&png_ptr, &info_ptr) Png_destroy_write_struct(&png_ptr, NULL) Info_ptr = png_create_info_struct(png_ptr) Png_ptr = png_create_write_struct(PNG_LIBPNG_VER_STRING, NULL, NULL, NULL) The following code should produce a 16-bit PNG, but instead produces 8-bit like this. I'm trying to write (using libpng) an 16-bit grayscale image where each point color equals to sum of its coordinates.
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