What is a good way to iterate a number through all the possible values of a mask?

Given a bitmask where the set bits describe where another number can be one or zero and the unset bits must be zero in开发者_运维问答 that number. What's a good way to iterate through all its possible values?

For example:

000 returns [000]
001 returns [000, 001]
010 returns [000, 010]
011 returns [000, 001, 010, 011]
100 returns [000, 100]
101 returns [000, 001, 100, 101]
110 returns [000, 010, 100, 110]
111 returns [000, 001, 010, 011, 100, 101, 110, 111]

The simplest way to do it would be to do it like this:

void f (int m) {
    int i;
    for (i = 0; i <= m; i++) {
        if (i == i & m)
            printf("%d\n", i);
    }
}

But this iterates through too many numbers. It should be on the order of 32 not 2**32.

There's a bit-twiddling trick for this (it's described in detail in Knuth's "The Art of Computer Programming" volume 4A §7.1.3; see p.150):

Given a mask mask and the current combination bits, you can generate the next combination with

bits = (bits - mask) & mask

...start at 0 and keep going until you get back to 0. (Use an unsigned integer type for portability; this won't work properly with signed integers on non-two's-complement machines. An unsigned integer is a better choice for a value being treated as a set of bits anyway.)

Example in C:

#include <stdio.h>

static void test(unsigned int mask)
{
    unsigned int bits = 0;

    printf("Testing %u:", mask);
    do {
        printf(" %u", bits);
        bits = (bits - mask) & mask;
    } while (bits != 0);
    printf("\n");
}

int main(void)
{
    unsigned int n;

    for (n = 0; n < 8; n++)
        test(n);
    return 0;
}

which gives:

Testing 0: 0
Testing 1: 0 1
Testing 2: 0 2
Testing 3: 0 1 2 3
Testing 4: 0 4
Testing 5: 0 1 4 5
Testing 6: 0 2 4 6
Testing 7: 0 1 2 3 4 5 6 7

(...and I agree that the answer for 000 should be [000]!)

First of all, it's unclear why 000 wouldn't return [000]. Is that a mistake?

Otherwise, given a mask value "m" and number "n" which meets the criterion (n & ~m)==0, I would suggest writing a formula to compute the next higher number. One such formula uses the operators "and", "or", "not", and "+", once each.

The trick by @Matthew is amazing. Here is a less tricky, but unfortunately also a less efficient, recursive version in Python:

def f(mask):
    if mask == "0":
        return ['0']
    elif mask == '1':
        return ['0', '1']
    else:
        bits1 = f(mask[1:])
        bits2 = []
        for b in bits1:
            bits2.append('0' + b)
            if mask[0] == '1':
                bits2.append('1' + b)
        return bits2

print f("101")  ===> ['000', '100', '001', '101']

You can do it brute-force. ;-) Ruby example:

require 'set'
set = Set.new
(0..n).each do |x|
  set << (x & n)
end

(where set is a set datatype, i.e., removes duplicates.)

Try this code:

def f (máscara):
    se máscara == "0":
        voltar ['0 ']
    elif máscara == '1 ':
        voltar ['0 ', '1']
    else:
        bits1 = f (máscara [1:])
        bits2 = []
        para b em bits1:
            bits2.append ('0 '+ b)
            se máscara [0] == '1 ':
                bits2.append ('1 '+ b)
        voltar bits2

print f ("101") ===> ['000 ', '100', '001 ', '101']

é interessante .