set integer value as bit mask

Question

What is simplest way to assign a bit mask to integer value? For example I want integer with first, third and forth bits = 1, other = 0.

Certainly I am looking for code, not for single value! And certainly there lot of possibilities, but I try to find simplest and most descriptive looking

Answer 1

This should do it:

int x = 0x0D;

And if you're lucky enough to use gcc and don't need to be portable:

int x = 0b1101;

Answer 2

Use the OR Bitwise operator (|) to combine bits:

#define FIRST_BIT (0x1)
#define SECOND_BIT (0x2)
#define THIRD_BIT (0x4)
#define FOURTH_BIT (0x8)
/* Increase by for each bit, *2 each time, 
   0x prefix means they're specified via a hex value */

int x = FIRST_BIT | THIRD_BIT | FOURTH_BIT;

And you can check if a bit is set using the AND Bitwise operator (&):

int isset = x&FIRST_BIT;

Answer 3

Here is an online bit converter, if you don't want to do the sum yourself:

http://www.binaryconvert.com/convert_unsigned_int.html

Just fill in the bits below (e.g. 1101), and compute the answer (e.g. 0x0000000D). Any capable calculator should be able to do the same ...

Answer 4

If you want your code to be readable in terms of the bit numbers, something like this may be useful:

#define b0  0x0001
#define b1  0x0002
#define b2  0x0004
#define b3  0x0008
#define b4  0x0010
:
#define b15 0x8000

int mask = b3|b2|b0;

But, after a while you should be able to tell which hex values relate to which bits and you won't need mnemonic tricks like that:

int mask = 0x000d;

Answer 5

int something = (0x00000001) & (0x00000001 * 2) until you get to where you want.

Answer 6

I think the best way to think (!) is to just index bits from 0, and then apply "to set the n:th bit, bitwise-OR with the value (1 << n)":

int first_third_and_fourth = (1 << 0) | (1 << 2) | (1 << 3);