Grabbing n bits from a byte

"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 = 00101b bottom = 42 & ((1 << 3) - 1) = 42 & (8 - 1) = 2 = 010b 

You could use bitfields for this. Bitfields are special structs where you can specify variables in bits.

typedef struct { unsigned char a:5; unsigned char b:3; } my_bit_t; unsigned char c = 0x42; my_bit_t * n = &c; int first = n->a; int sec = n->b; 

Bit fields are described in more detail at http://www.cs.cf.ac.uk/Dave/C/node13.html#SECTION001320000000000000000

The charm of bit fields is, that you do not have to deal with shift operators etc. The notation is quite easy. As always with manipulating bits there is a portability issue.

int x = (number >> 3) & 0x1f;

will give you an integer where the last 5 bits are the 8-4 bits of number and zeros in the other bits.

Similarly,

int y = number & 0x7;

will give you an integer with the last 3 bits set the last 3 bits of number and the zeros in the rest.

just get rid of the 8* in your code.

int input = 42; int high3 = input >> 5; int low5 = input & (32 - 1); // 32 = 2^5 bool isBit3On = input & 4; // 4 = 2^(3-1) 

I think there is a much simpler way to do what you want. You only need the & "and" operator.

Basically, if you want the 5 most significant bits, all you have to do is add them up. 128+64+32+16+8 = 248 or the 3 least significant bits would be 4+2+1=7. Then use the and operator ( & ) with your variable and the total to assign those bits to a new variable.

int main() { unsigned char c = 42; //sum of 3rd bit + 5th bit + 7th bit int first5 = c & 248; //sum of bits 1-5 int last3 = c & 7; // sum of bits 6-8 cout << "first 5 " << first5 << " last 3 " << last3 << endl; cout << "OR 5 most significant bits " << first5 << endl; cout << "and 3 least significant bits " << last3 <<endl; return 0; } 

But, you can also use this to grab one individual bit or any combination of bits.

For instance...

int threenfive = c&40; //would extract the 3rd bit and 5th bit 

because the 3rd bit = 32 and the 5th = 8 and the sum of the two = 40.