How do i get the lower 8 bits of int?

Question

Lets say I have an int variable n = 8. On most machines this will be a 32 bit value. How can I only get the lower 8 bits (lowest byte) of this in binary? Also how can i acce

Answer 1

Use bitwise arithmetic to mask off the lowest 8 bits:

unsigned char c = (x & 0xFF);

To access the nth lowest bit, the equation is (x & (1 << n)) (n of zero indicates the least significant bit). A result of zero indicates the bit is clear, and non-zero indicates the bit is set.

Answer 2

You can test if a particular bit is set in a number using << and &, ie:

if (num & (1<<3)) ...

will test if the fourth bit is set or not.

Similarly, you can extract just the lowest 8 bits (as an integer) by using & with a number which only has the lowest 8 bits set, ie num & 255 or num & 0xFF (in hexadecimal).

Answer 3

The best way is to use the bit logical operator & with the proper value.

So for the lower 8 bits:

n & 0xFF; /* 0xFF == all the lower 8 bits set */

Or as a general rule:

n & ((1<<8)-1) /* generate 0x100 then subtract 1, thus 0xFF */

You can combine with the bit shift operator to get a specific bit:

(n & (1<<3))>>3; 
  /* will give the value of the 3rd bit - note the >>3 is just to make the value either 0, or 1, not 0 or non-0 */

Answer 4

unsigned n = 8;
unsigned low8bits = n & 0xFF;

Note a few things:

1. For bitwise operations, always use the unsigned types
2. Bits can be extracted from numbers using binary masking with the & operator
3. To access the low 8 bits the mask is 0xFF because in binary it has its low 8 bits turned on and the rest 0
4. The low 8 bits of the number 8 are... 8 (think about it for a moment)

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To access a certain bit of a number, say the kth bit:

unsigned n = ...;
unsigned kthbit = (1 << k) & n;

Now, kthbit will be 0 if the kth bit of n is 0, and some positive number (2**k) if the kth bit of n is 1.