C how to roll bytes

Question

This is just something I wanted to do for fun, no real practical use of it.

Say for example I enter 0xDEADAAAA

I want to be able to transform the value into 0xDDEAAAAA

Meaning the 4 most significant bytes shift over one byte. And the 4th most significant byte now becomes the first most significant byte.

Kind of like a circular shift of the first 4 bytes.

For now I'm doing

value = value >> ((sizeof(unsigned int) * 8) / 2);

This will shift all bytes over one byte. But how would I do the circular roll of the 4th most significant byte into the MSB?

A string has no "most significant byte". And if that is supposed to be an integer: it seem to only have 4 bytes. It is not clear what you mean. See How to Ask, provide a minimal reproducible example. And search for "c operators", if that is not too much asked. — too honest for this site
– too honest for this site, Commented Sep 30, 2016 at 0:15
See Bit Twiddling Hacks. You need to rotate 4 high-order bytes while leaving the low-ones unchanged (though with the value AAAA in the low-order bytes, it will be hard to spot mishandled changes that preserve byte values; use 0xDEAD0123 or something else with distinct bytes in the low-order part of the value. Also, you seem to be mixing up bytes with nybbles; each hex digit is a nybble (aka nibble), and it appears you want to deal with the 4 most significant nybbles, not bytes. — Jonathan Leffler
– Jonathan Leffler, Commented Sep 30, 2016 at 0:18
It's an unsigned integer not a string. I'm just representing it in hexadecimal for readability. — momonkey7
– momonkey7, Commented Sep 30, 2016 at 0:18
Ah you're right yes nibbles not bytes then. Because a nibble is 4 binary digits correct? And a byte is 8? — momonkey7
– momonkey7, Commented Sep 30, 2016 at 0:28
Yes; a nybble is 4 bits and a byte is 8 bits on most systems these days. (The exceptions are few and far between and I've never programmed on any of them.) — Jonathan Leffler
– Jonathan Leffler, Commented Sep 30, 2016 at 0:34

Stack Exchange Broke The Law · Accepted Answer · 2016-09-30 00:38:16Z

You can split the number into three parts and process them separately, then combine them.

The part that is unchanged is value & 0x0000FFFF.
The part that you want to shift right 4 bits is value & 0xFFF00000.
The part that you want to shift left 12 bits is value & 0x000F0000.

So do all of those things, then combine the results with |:

value = (value & 0x0000FFFF) | ((value & 0x000F0000) << 12) | ((value & 0xFFF00000) >> 4);

(note, here I've lined up all the masks so it's easy to see, but obviously you don't have to do that)

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