Question

Given an unsigned int A (32 bit), and another unsigned int B, where B’s binary form denotes the 10 “least reliable” bits of A, what is the fastest way to expand all 1024 potential values of A? I’m looking to do this in C.

E.g uint B is guaranteed to always have 10 1’s and 22 0’s in it’s binary form (10 least reliable bits).

For example, let’s say

A = 2323409845  
B = 1145324694

Their binary representations are:

a=10001010011111000110101110110101

b=01000100010001000100010010010110

B denotes the 10 least reliable bits of A. So each bit that is set to 1 in B denotes an unreliable bit in A.

I would like to calculate all 1024 possible values created by toggling any of those 10 bits in A.

Answer 1

You can iterate through the 1024 different settings of the bits in b like so:

unsigned long b = 1145324694;
unsigned long c;

c = 0;
do {
    printf("%#.8lx\n", c & b);
    c = (c | ~b) + 1;
} while (c);

To use these to modify a you can just use XOR:

unsigned long a = 2323409845;
unsigned long b = 1145324694;
unsigned long c;

c = 0;
do {
    printf("%#.8lx\n", a ^ (c & b));
    c = (c | ~b) + 1;
} while (c);

This method has the advantages that you don’t need to precalculate any tables, and you don’t need to hardcode the 1024 – it will loop based entirely on the number of 1 bits in b.

It’s also a relatively simple matter to parallelise this algorithm using integer vector instructions.