Converting two 32-bit ints to a u_int64_t...

I have a function that retrieves the high and low order bits of the CPUs cycle-counter (Using GCC asm):

Code:

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void access_counter(unsigned *hi, unsigned *lo) {
        asm("rdtsc; movl %%edx,%0; movl %%eax,%1"
        : "=r" (*hi), "=r" (*lo)
        :
        : "%edx", "%eax");
}

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... And I am writing a function that returns a u_int64_t based on the high and low order bits of access_counter():

Code:

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u_int64_t get_cycles(void) {
        unsigned hi, lo;
        u_int64_t hi_shift;

        access_counter(&hi, &lo);

        hi_shift = hi;

        return (hi_shift << 32) + lo;
}

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... This seems like the correct way of doing this. The only problem is that I cannot think of a way to check if it is correct. So I was hoping someone could look this over and see if this correctly returns the value of the cycle counter as a u_int64_t.

Thanks

That should work, but a union would be easier.

Quote:

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Originally Posted by iMalc

That should work, but a union would be easier.

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I've never heard of unions in C. How would that work in this context?

Also.. How could I print a u_int64_t to stdout? %lld doesn't seem to work.

It's not u_int64_t, it's uint64_t. [edit]u_int64_t exists on some platforms.[/edit]

lld is the conversion specification for a long long signed int.

Use PRIu64 in inttypes.h:

Code:

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uint64_t a=123;
printf("It is: %" PRIu64 "\n",a);

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see 7.8.1 of the C standard.

Quote:

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Originally Posted by elfjuice

I've never heard of unions in C. How would that work in this context?

Also.. How could I print a u_int64_t to stdout? %lld doesn't seem to work.

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This is a slightly simplified version of LARGE_INTEGER from WinNT.h in VS2005 Express:

Code:

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typedef union _LARGE_INTEGER {
    struct {
        DWORD LowPart;
        LONG HighPart;
    };
    LONGLONG QuadPart;
} LARGE_INTEGER;

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You can use it like this:

Code:

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u_int64_t get_cycles(void) {
        LARGE_INTEGER l;

        access_counter(&l.HighPart, &l.LowPart);

        return l.QuadPart;
}

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Quote:

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Originally Posted by robwhit

It's not u_int64_t, it's uint64_t.

llu is the conversion specification for a long long unsigned int.

Use PRIu64 in inttypes.h:

Code:

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uint64_t a=123;
printf("It is: " PRIu64 "\n",a);

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see 7.8.1 of the C standard.

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It is u_int64_t as defined in sys/types.h (unix).

Thanks for the help. But if my shift method works correctly then I will leave it how it is.

Thanks again.

Yes, I had been doing some research on that and it appears that I'll have to eat my words.

My post has been edited to reflect some changes.

I can't find anywhere that u_int64_t is anything other than an unsigned long long int, so I guess you could use %llu, which coincidentally goes right against what I originally had my post saying.[edit]ok that's not true either... lemme do some more research...[/edit]

Yes! I found it. Yes, I was looking all that time. :D

u_int64_t i;
printf ("%ju", (uintmax_t) i);

http://www.daemonnews.org/mailinglis.../msg00397.html

This will work for any integer type under C99, regardless of what the typedef is.

Quote:

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Originally Posted by robwhit

Yes! I found it. Yes, I was looking all that time. :D

u_int64_t i;
printf ("%ju", (uintmax_t) i);

http://www.daemonnews.org/mailinglis.../msg00397.html

This will work for any integer type under C99, regardless of what the typedef is.

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Awesome. That helps.

Thanks