128-bit struct or 2 64-bit records for performance and readibility

Question

I am sorting 128-bit records in a large file (10s of GB). Algorithm works fine and currently using uint64_t with two of these equal to one record. This is on a modern 64-bit system.

In the interest of code readability, am wondering if create a struct like typedef struct u128t {uint64_t hi, uint64_t} u128t can be used in place of the two 64-bit records. This would make array indexing and assignment cleaner but I would have to implement a comparator function.

Is this a portable solution and should I expect this to run at the same speed as the original implementation?

Further reading:

http://gcc.gnu.org/onlinedocs/gcc/_005f_005fint128.html - for 128-bit systems
http://en.wikipedia.org/wiki/C_data_types - c types
Copy struct to struct in C - copy structs

With an inlined comparator and a compiler that is not completely braindead there should be no noticeable performance decrease. The only way to be sure is of course to try it and measure. — ComicSansMS, Sep 19 '13 at 14:09
As one uses `struct u128t` beyond simple compares, it may be useful to order the fields `uint64_t hi` and `uint64_t lo` with the machine's endian-ness. Makes other 128 bit operations easier. — chux - Reinstate Monica, Sep 19 '13 at 14:52

score 5 · Accepted Answer · answered Sep 19 '13 at 13:48

5

Using struct is fine.

I would do something like this:

#if COMPILER_WHICH_SUPPORTS_128_BIT_TYPE
    typedef __uint128 u128t;
    #define COMPARE(a, b) ((a) == (b))
#else
    typedef struct {
        uint64_t hi;
        uint64_t lo;
    } u128t;
    #define COMPARE(a, b)   MyCompareFunction((a), (b))
#endif

answered Sep 19 '13 at 13:48

user694733

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+1 for COMPILER_WHICH_SUPPORTS_128_BIT_TYPE. This solution is both readable and (I guess) ideal for performance. – johan d Sep 19 '13 at 14:01

128-bit struct or 2 64-bit records for performance and readibility

1 Answers1