There is some discussion in the Rust Reddit, as Rust since some time has some benchmarks on the Computer Game site:
http://www.reddit.com/r/rust/comments/1xcq1q/c_gcc_vs_rust_wtf/

The source code of the Rust benchmarks is probably here:
https://github.com/mozilla/rust/tree/master/src/test/bench

They are mostly discussing about the very slow big int implementation.

A short D implementation of the the pidigits benchmark that uses Phobos BigInts (this is not meant to compete with the entries that use GMP. A D entry using GMP is possible), adapted from the C entry by the good Ledrug:

http://dpaste.dzfl.pl/821527e71343

I have not compared them, but I think it's 3-4 times slower than the best entry (that uses GMP).

I suspect that if Phobos BigInts gets mutable buffers (https://d.puremagic.com/issues/show_bug.cgi?id=7013 ), this code could be modified to be rather faster.

Bye,
bearophile

On Monday, 10 February 2014 at 22:19:19 UTC, bearophile wrote:

> http://dpaste.dzfl.pl/821527e71343

Your paste has expired / no longer there.... but the subject has come up again...

http://www.wilfred.me.uk/blog/2014/10/20/the-fastest-bigint-in-the-west/

https://lobste.rs/s/sfie8j/the_fastest_bigint_in_the_west

I thought to take a poke at it from this point of view.

Ruby "wins" the game for shortest code... I wondered whether D could score high both on the terse/readable and speed categories.

Do you still have your implementation hanging around?

John Carter:

> Your paste has expired / no longer there.... but the subject has come up again...
> ...
> Do you still have your implementation hanging around?

I think it was this. *Untested*:


void main(string[] args) {
    import core.stdc.stdio, std.bigint, core.stdc.stdlib;

    immutable n = (args.length == 2) ? args[1].ptr.atoi : 100;
    BigInt acc, den = 1, num = 1;

    for (uint i, k; i < n; ) {
        immutable k2 = ++k * 2U + 1U;
        acc = (acc + num * 2U) * k2;
        den *= k2;
        num *= k;
        if (num > acc)
            continue;

        immutable d = ((num * 3 + acc) / den) % 10U;
        if (d != ((num * 4 + acc) / den) % 10U)
            continue;

        putchar('0' + d);
        if (++i % 10 == 0)
            printf("\t:%u\n", i);
        acc = (acc - den * d) * 10U;
        num *= 10U;
    }
}

Bye,
bearophile