Set n bits mask starting with k position in unsigned int

Question

suppose I want to apply mask 11d to unsigned int starting at position k (in that unsigned int). how can this be implemented?

I wrote a function that implements this, maybe it will be useful to someone

#include <stdio.h>
#include <limits.h>

#define BITS_PER_INT (CHAR_BIT * sizeof(unsigned int))

unsigned int set_bitmask_from_k_position(unsigned int x, int k, unsigned int pattern)
{
    int nbits;
    unsigned int tmp;
    for (nbits = 0, tmp = pattern; tmp > 0; tmp /= 2, nbits++); 
    unsigned int mask = 0;
    mask|=mask | ((~0u << (k+nbits-1))|(~0u >> (BITS_PER_INT-k+1)));
    x &= mask;
    mask = pattern << (k - 1);
    x |= mask;
    return x;
}
int main(void)
{
    printf("0x%X\n", set_bitmask_from_k_position(0x12345678u, 12, 11));
}

I answered my own question myself, thanks to everyone who helped and even those who downgraded my question, I did not find such a question on stackowerflow, so I asked

Like this: unsigned int mask =11; unsigned int x = 0x12345678; mask <<= 15; x |= mask — Ivan Ivanovich, Feb 17 '21 at 22:34
`11` decimal? Or binary? What is the desired result? You also seem to know how to do it, so what is the question? — Eugene Sh., Feb 17 '21 at 22:35
you first need to zero out these bits. Like in `x &= ~(0xF << 12)` — Eugene Sh., Feb 17 '21 at 22:38
`unsigned int mask =(11 << 12); unsigned int x = 0x12345678; x &= ~(0xF << 12); x |= mask;` ? — isrnick, Feb 17 '21 at 22:41
Maybe use [bit fields](https://stackoverflow.com/q/24933242/10077)? — Fred Larson, Feb 17 '21 at 22:52
the task is not to lose the value of x, but only to modify its 4 bytes starting from position 12 to position 15, modify it's 4 bytes to value 11d — Ivan Ivanovich, Feb 17 '21 at 22:53
To set those 4 bits to 11 decimal, you need to and `&` away the old 4 bits and or `|` in the new 4 bits, more than likely by shifting `<<` the 11 value up to the 12th position first. Basically what @EugeneSh. and others are suggesting. Having the C code of your attempt would answer a lot of questions... — Michael Dorgan, Feb 17 '21 at 23:07
See [Bit Twiddling Hacks](http://graphics.stanford.edu/~seander/bithacks.html). — Jonathan Leffler, Feb 17 '21 at 23:23
See also [Bit mask in C](https://stackoverflow.com/q/316488/15168) for ways to set _m_ bits to 1 and _n_ bits to 0 in the less significant bit positions. This would be useful for zeroing the bits that need to be set to 11 (decimal, 0b1011 binary, 0x0B hex, 013 octal) before adding those bits back in. — Jonathan Leffler, Feb 17 '21 at 23:28
i wrote the function you are looking for, see updated question — Ivan Ivanovich, Feb 18 '21 at 08:13

score 2 · Accepted Answer · answered Feb 17 '21 at 23:07

2

Make a bit mask for the number of bits from 12 to 15 (inclusive) by generating the value 2^15−12+1−1:

unsigned mask = (1u << (15-12+1)) - 1u;

Move the mask to bits 12 to 15:

mask <<= 12;

Use the mask to turn off bits 12 to 15 in x:

x &= ~mask;

Put 11 in those bits:

x |= 11 << 12;

Do it in one step:

x = x & ~( ((1u << (15-12+1)) - 1u) << 12 ) | (11 < 12);

I might also write it as:

x = x & ~( (2u<<15) - (1u<<12) ) | (11 < 12);

because the 2u<<15 combines the +1 into the shift and avoids a problem with C semantics if we want to use a field reaching the highest bit in an unsigned.

answered Feb 17 '21 at 23:07

Eric Postpischil

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I really hope his unsigned ints aren't 16-bits in this day and age... – Michael Dorgan Feb 17 '21 at 23:11
thanks for the help. I wrote a function according to your prompts, it is in my post – Ivan Ivanovich Feb 18 '21 at 08:46

Set n bits mask starting with k position in unsigned int

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