Is there any way to get 64-bit time_t in 32-bit programs in Linux?

Question

On Windows I can call:

_time32(__time32_t); // to get 32-bit time_t
_time64(__time64_t); // to get 64-bit time_t

(both in 32 and 64-bit programs)

Is there any way do this in Linux (compiling with GCC)?

Answer 1

Apparently, no it's not possible. For starters, there is only one time() function in Linux, no time32() or time64().

After searching for a while, I can see that it's not libc's fault, but the culprit is actually the kernel.

In order for libc to fetch the current time, it need to execute a system call for it: ^(Source)

time_t time (t) time_t *t;
{
    // ...
    INTERNAL_SYSCALL_DECL (err);
    time_t res = INTERNAL_SYSCALL (time, err, 1, NULL);
    // ...
    return res;
}

The system call is defined as: ^(Source)

SYSCALL_DEFINE1(time, time_t __user *, tloc)
{
    time_t i = get_seconds();
    // ...
    return i;
}

The function get_seconds() returns an unsigned long, like so: ^(Source)

unsigned long get_seconds(void)
{
    struct timekeeper *tk = &timekeeper;

    return tk->xtime_sec;
}

And timekeeper.xtime_sec is actually 64-bit: ^(Source)

struct timekeeper {
    // ...
    /* Current CLOCK_REALTIME time in seconds */
    u64                     xtime_sec;
    // ...
}

Now, if you know your C, you know that the size of unsigned long is actually implementation-dependant. On my 64-bit machine here, it's 64-bit; but on my 32-bit machine here, it's 32-bit. It possibly could be 64-bit on some 32-bit implementation, but there's no guarantee.

On the other hand, u64 is always 64-bit, so at the very base, the kernel keeps track of the time in a 64-bit type. Why it then proceeds to return this as an unsigned long, which is not guaranteed to be 64-bit long, is beyond me.

In the end, even if libc's would force time_t to hold a 64-bit value, it wouldn't change a thing.

You could tie your application deeply into the kernel, but I don't think it's even worth it.