How do these image bit conversions work??

So this is a basic question on image conversion algorithms.

If I load an image.

 img = imread('peppers.png');
 img = rgb2gray(img);

and then convert it back forth from 8 bit to 16 bit several times;

 img = im2uint16(img);
 img =im2uint8(img);

for like 10 times or so. The final resulting image seems exactly the same. That seems a bit odd to me, shouldn't there be some rounding errors when converting from 16-bit to 8-bit? That information should have been lost right? And shouldn't each iteration of this conversion amplify the error?

I am using MATLAB 2017b so its entirely possible that older versions of matlab may not do this...

Answer 1

[Walter Roberson]

im2uint16() notices that the input is uint8, and multiplies the input by a constant number to get the 16 bit version. The number it multiplies by is 257. This is the same thing as duplicating the bytes -- for example byte 0xD8 gets changed to 0xD8D8

im2uint8() applied to a uint16 image works by just dropping the last byte.

You can see that a cycle of these involves duplicating bytes and dropping them, which is going to leave you with the original bytes.

[Nick]

Oh thats right, I was multiplying things by 256 and wondering why everything was off.