Bits operations implementations

Question

I have to return 1 if bit i in x is set, 0 otherwise in is_set function. I got stuck here. Have no idea what to do next...Any ideas?? Any help will appreciated....

#include <string.h>
#include <stdio.h>

const char *to_binary(unsigned int x) {
    static char bits[17];
    bits[0] = '\0';
    unsigned int z;
    for (z = 1 << 15; z > 0; z >>= 1) {
        strcat(bits, (x & z) ? "1" : "0");
    }
    return bits;

Um, you virtually included the answer in the code you posted ... consider `x & z`, which is performed 16 times -- what do you think that does on each of those iterations? — Jim Balter, Feb 06 '14 at 05:54
`x & z` ... what does z contain? More specifically: when z is 1 << 15, what does the expression do? Don't just say "bitwise and" ... what does it *do*? — Jim Balter, Feb 06 '14 at 05:57
Lol sorry I am beginner in this....I dont understand what u r asking.... :( — user3247903, Feb 06 '14 at 06:00
What does x & z do when z is 1 << 15? It tests bit 15 of x. What does x & z do when z is 1 << 9? It tests bit 9 of x. What does `(x & (1 << bit))` do? — Jim Balter, Feb 06 '14 at 06:03

Jim Balter · Accepted Answer · 2014-02-06T06:21:30.497

2

short is_set(unsigned short x, int bit) {
    return x & (1 << bit) ? 1 : 0; 
}

Alternatively,

short is_set(unsigned short x, int bit) {
    return (x >> bit) & 1; 
}

edited Feb 06 '14 at 06:21

answered Feb 06 '14 at 06:05

Jim Balter

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I feel `return x & (1 << bit)` is enough. – Jeyaram Feb 06 '14 at 06:14
1

@Jeyaram You feel wrong ... the requirement is to return 1 if the bit is set, not `1 << bit`. If I were interviewing you, I would blackball you for that. OTOH, `(x >> bit) & 1` is ok, but more obscure. – Jim Balter Feb 06 '14 at 06:15

Bits operations implementations

1 Answers1