Write a function setbits(x,p,n,y) that returns x with the n bits that begin at position p set to the rightmost n bits of y, leaving the other bits unchanged.

#include <stdio.h>

unsigned setbits(unsigned x, int p, int n,unsigned y);

int main(void){
    unsigned result=setbits(0, 4, 3, ~0);
    return 0;

unsigned setbits(unsigned x, int p, int n,unsigned y){
    unsigned ybits = y & ~(~0 << n);
    unsigned helper = ~(~0 << n);
    unsigned result = (x | (helper << (p+1-n))) & ((ybits << (p+1-n)) | (~(helper << (p+1-n))));
    return result;



  • \$\begingroup\$ Are the 3 bits that start at bit 4 bits 4,5,6 or bits 4,3,2? \$\endgroup\$ – JS1 Nov 7 '16 at 6:17

Your example, which uses x = 0 and y = ~0, is not a very useful demonstration, as those degenerate cases would tend to let bugs remain undetected.

You have some common expressions that reappear several times, which should be given variable names:

  • (p+1-n) appears three times in result. I would name it shift.
  • ~(~0 << n) appears in ybits, but it is the same as helper. I would name it mask.

    I would write it as (1 << n) - 1 instead. It's probably not any more or less efficient than ~(~0 << n). However, you use bitwise negation a lot. When you write (~(helper << (p+1-n))), that's a negation of a negation of a negation, which I find headache-inducing.

Conversely, you gave variable names to ybits and result, neither of which provide any benefit.

Your strategy is, in summary:

(x | ForceRelevantBitsOn) & ((ybits << ShiftToRelevantPlace) | OnesEverywhereElse)

However, that strategy requires you to construct OnesEverywhereElse. I recommend the following simpler technique instead:

(x & ForceRelevantBitsOff) | (ybits << ShiftToRelevantPlace)

Here's how I would write it:

unsigned setbits(unsigned x, int p, int n, unsigned y) {
    int shift = p - n + 1;
    unsigned mask = (1 << n) - 1;
    return (x & ~(mask << shift)) | ((y & mask) << shift);

The last line could be replaced with this expression, which might be more efficient (using five operations instead of six), but is more cryptic:

    return ((((x >> shift) ^ y) & mask) << shift) ^ x;

That works by determining which bits in x need to be toggled.

| improve this answer | |
  • \$\begingroup\$ Nice improvement.Just there is a part i do not understand.You said that ~(~0 << n) is equal to (1 << n) - 1 but why ? Firstly 1 is not equal to ~0 right ? And secondly why do you substract it from 1 ? \$\endgroup\$ – Silidrone Nov 8 '16 at 17:41

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