Grabbing n bits from a byte

Integers are represented inside a machine as a sequence of bits; fortunately for us humans, programming languages provide a mechanism to show us these numbers in decimal (or hexadecimal), but that does not alter their internal representation.

You should revise the bitwise operators &, |, ^ and ~ as well as the shift operators << and >>, which will help you understand how to solve problems like this.

The last 3 bits of the integer are:

x & 0x7

The five bits starting from the eight-last bit are:

x >> 3    // all but the last three bits  &  0x1F // the last five bits.

"grabbing" parts of an integer type in C works like this:

1. You shift the bits you want to the lowest position.
2. You use & to mask the bits you want - ones means "copy this bit", zeros mean "ignore"

So, in you example. Let's say we have a number int x = 42;

first 5 bits:

(x >> 3) & ((1 << 5)-1);

or

(x >> 3) & 31;

To fetch the lower three bits:

(x >> 0) & ((1 << 3)-1)

or:

x & 7;

Say you want hi bits from the top, and lo bits from the bottom. (5 and 3 in your example)

top = (n >> lo) & ((1 << hi) - 1)bottom = n & ((1 << lo) - 1)

Explanation:

For the top, first get rid of the lower bits (shift right), then mask the remaining with an "all ones" mask (if you have a binary number like 0010000, subtracting one results 0001111 - the same number of 1s as you had 0-s in the original number).

For the bottom it's the same, just don't have to care with the initial shifting.

top = (42 >> 3) & ((1 << 5) - 1) = 5 & (32 - 1) = 5 = 00101bbottom = 42 & ((1 << 3) - 1) = 42 & (8 - 1) = 2 = 010b