Most correct way for bit checking in C - c

Question

I reading a book "Linux Kernel. Development. Third Edition." by Robert Love.
And in softirq section it's comment next piece of code:
u32 pending;
pending = local_softirq_pending();
if (pending) {
struct softirq_action *h;
/* reset the pending bitmask */
set_softirq_pending(0);
h = softirq_vec;
do {
if (pending & 1) /* STEP 4 */
h->action(h);
h++;
pending >>= 1;
} while (pending);
}
He discribe step by step what happens and most unclear step for me is 4:
If the first bit in pending is set, h->action(h) is called
I have following code to check if bit set like in book:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define BIT_SET(n) ((1) << (n))
#define BIT_CLEAR(n) ~((1) << (n))
int main(void)
{
unsigned char bits = 0x0;
bits |= BIT_SET(1);
if (bits & (1 << 1))
printf("TEST CHECK SET 1\n");
if (bits & 1)
printf("TEST CHECK SET 2\n");
bits &= BIT_CLEAR(1);
if (!(bits >> 1) & 1UL)
printf("BITS UNSET\n");
return 0;
}
compiled with:
gcc main.c -O0 -Wall -Wextra -Werror.
I always check if bit set with this one:
if (bits & (1 << n))
And my code make output like this:
TEST CHECK SET 1
BITS UNSET
Why if (bits & 1) statment is not work?:
So from several option which should I use and what exactly last one check?
if (bit & (1 << n))
if ((bit >> n) & 1)
if (bit & n)

Answer 1

I always check if bit set with this one: if (bits & (1 << n))
This checks n-th bit in place; the code from the book, however, shifts the bit into the least-significant-bit position prior to making the check with bits & 1. In other words, by the time the code reaches the if (bits & 1) the value of bits has already been shifted such that the bit of interest is in 1-s position.
This is similar to your other check
if ((bit >> n) & 1)
except (bit >> n) part is done by performing bit >>= 1 operation n times in a loop.
Note that in order for this code to work correctly bit must be unsigned.
So from several option which should I use and what exactly last one check?
You misinterpreted the last check: it's not checking bit n, it's checking bit against the entire bit pattern of n's binary representation.