How to for() loop with hex numbers?

[monk64]

I know, I know, inside the software, a number is a number. But here's my problem...

I want to do something like this:

for (i=0; i<=63; i++) {
      if ( ( flags & (1 << i ) ) != 0 ) 
           printf ("bit %i is set\n", i);
     else
           printf ("bit %i is not set\n", i);
}

That doesn't work, though, because for example if i is 49, that's interpreted as hex and it tries to get info for bit 73 (49 in hex is 73).

Of course, I could write 0x31 if I knew when I was writing the code that I wanted to talk to bit 49, but if I'm in a loop, or have an algorithm that might determine which bit I want to talk do, how do I say "interpret this number as hex"?

In other words, how can I write

      if ( ( flags & (1 << somevar ) ) != 0 )

such that somevar is in interpreted as hex? A lot of the examples I've googled assume we're dealing with an 8-bit char, which is a moot point because 1-8 is the same in either decimal or hex ;-)

[cyberfish]

It will work. Nothing will be interpreted in hex. Like you said, a number is just a number.

1 << 10

will shift left by 10 (decimal) bits.

[monk64]

It will work. Nothing will be interpreted in hex. Like you said, a number is just a number.

1 << 10

will shift left by 10 (decimal) bits.

I thought so, too, but...well, perhaps I have a different problem. I am writing code on 32-bit linux. According to all docs I can find, unsigned long long ints are guaranteed to be 64 bits.

$ cat example.c
#include <stdio.h>

int main (void) {
        unsigned long long int flags;
        int i;

        flags = 0;

        flags |= (1 << 2);
        flags |= (1 << 21);
        flags |= (1 << 49);
        flags |= (1 << 63);

        for (i =0; i<=63; i++) {
                if ( ( flags & (1 << i ) ) != 0 )
                        printf ("bit %i is set\n",i);
        }

}

$ gcc -o example example.c
example.c: In function ‘main’:
example.c:11: warning: left shift count >= width of type
example.c:12: warning: left shift count >= width of type

$ ./example
bit 2 is set
bit 21 is set
bit 34 is set   <- incorrect
bit 53 is set   <- incorrect

I'd assumed it was a hex notation problem...perhaps it's something else?

[tabstop]

Easy way to check:

print("There are %d bits in an unsigned long long.\n", 8*sizeof(unsigned long long int));

EDIT EDIT EDIT: Although, the catch is, "1" is not an unsigned long long. Your shift should be

1ULL << 49

[cyberfish]

(1 << 63)

Type of 1 is int, type of 63 is int, so the type of the result will also be int (truncated).

Then, it will be sign or zero extended to 64-bits and stored into flag.

Try
(1LL << 63)

[monk64]

Easy way to check:

print("There are %d bits in an unsigned long long.\n", 8*sizeof(unsigned long long int));

EDIT EDIT EDIT: Although, the catch is, "1" is not an unsigned long long. Your shift should be

1ULL << 49

Bravissimo! That is it. Replacing "1" with "1ULL" fixed it. Makes complete sense.

Thanks much - this was quite educational.