We can say that clang is like the young Gauss :)

http://mathandmultimedia.com/2010/09/15/sum-first-n-positive...

Wait, that contains no loop. Is clang using mathematical rules for series summation? Awesome!

It gets even better after playing with the code a bit: https://godbolt.org/g/kUMCpl

magic constants galore... where does 3435973837 come from? ;)

First it normalizes to

    if (num <= 0) {
    	return 0;
    }
  
    int a = 0;
    int loop_n = (unsigned)(num - 1) / 5;
    for (int x = 0; x <= loop_n; x++) {
        a += x * 5 - 1;
    }
    return a;

Division by 5 is done by multiplying by the modular inverse, 3435973837.

From there you just have a summation, which has two components.

    ((loop_n + 1) * loop_n / 2) * 5    // 5x
    - loop_n - 1                       // -1

The assembly between these the manual and automatic methods is slightly different, but the difference is fairly trivial and opaque.

It's 0xCCCCCCCD in hex, which looks a lot more structured at least. Though how exactly this makes sense is still unclear :-)

[edit] and playing around with the increment of x it keeps throwing in some magical values which are a repeating pattern in hex, with the LSB off by one usually...

I thought 0xCCCCCCCD was a very nice number so I asked myself why it would come out like that. I wondered whether

0.CCCCCCCC.... = 4/5 had anything to do with it.

Using hexadecimal we have

    5 * 3 = 10 - 1, so 
    5 * C = 40 - 4 so
    5 * (1 + C + C0 + C00 + C000...) = 5 - 4 + 40 - 40 + 400... where the last term is = 0 mod 2^32
         -----
           D

This relates to

    0.CCCCCC.... = 4 / 5

Because in order for that to come out right, multiplying 0.CCCC... with 5 has to have all intermediate terms cancel, leaving only a 4.

So why would it take its digit from 4 / 5 and not 1 / 5? Because it takes x=4 to make the 5-x subtraction in 5 - x + x0 - x0 + x00 to come out to 1.

So this gives us a general rule for inversion (for numbers less than 16). Take the hexadecimal expansion of (n-1) / n, truncate to get enough hex chars, add 1 to the least significant hex digit.

I.e. the inversion of 3 would be 0xAAAAAAAB

Well if A=1 (I know it’s not), D is 5-1. I’ve noticed that whatever you change five to fits this pattern. Ex: +=3 in the code gives you AAAAAAAB, and B would be (3-1) if A=1. I