jg | A CLI tool to generate JSON from the command-line | Command Line Interface library

This happens because jg and jl treat the outcome of cmp as if the two operands were signed numbers.

7Ah and 8Dh represent signed numbers +122 and -115, respectively. Obviously, the latter is the smallest.

What you need is unsigned comparison. Use instructions ja and jb instead.

If you're trying to copy the SQL exactly I think you missed the 'W' in the cfset at the top:

Outputs:

Week: W9 (This is the value of the WEEK_2 variable)

The GraphTraversal object is only a "plan" to be carried out. To have it take effect, you need a closing method like next, toList, etc., like you did for the count.

The confusion probably arose from the fact that the gremlin console automatically keeps nexting the traversal a configured number of times.

I do decoding like below:

Your assembly code seems correct! It does contain a lot of redundant instructions though.

Your solution that works from 5 numbers pushed on the stack deserves a loop:

It looks like GCC's naïveté about store-forwarding stalls is hurting its auto-vectorization strategy here. See also Store forwarding by example for some practical benchmarks on Intel with hardware performance counters, and What are the costs of failed store-to-load forwarding on x86? Also Agner Fog's x86 optimization guides.

(gcc -O3 enables -ftree-vectorize and a few other options not included by -O2, e.g. if-conversion to branchless cmov, which is another way -O3 can hurt with data patterns GCC didn't expect. By comparison, Clang enables auto-vectorization even at -O2, although some of its optimizations are still only on at -O3.)

It's doing 64-bit loads (and branching to store or not) on pairs of ints. This means, if we swapped the last iteration, this load comes half from that store, half from fresh memory, so we get a store-forwarding stall after every swap. But bubble sort often has long chains of swapping every iteration as an element bubbles far, so this is really bad.

(Bubble sort is bad in general, especially if implemented naively without keeping the previous iteration's second element around in a register. It can be interesting to analyze the asm details of exactly why it sucks, so it is fair enough for wanting to try.)

Anyway, this is pretty clearly an anti-optimization you should report on GCC Bugzilla with the "missed-optimization" keyword. Scalar loads are cheap, and store-forwarding stalls are costly. (Can modern x86 implementations store-forward from more than one prior store? no, nor can microarchitectures other than in-order Atom efficiently load when it partially overlaps with one previous store, and partially from data that has to come from the L1d cache.)

Even better would be to keep buf[x+1] in a register and use it as buf[x] in the next iteration, avoiding a store and load. (Like good hand-written asm bubble sort examples, a few of which exist on Stack Overflow.)

If it wasn't for the store-forwarding stalls (which AFAIK GCC doesn't know about in its cost model), this strategy might be about break-even. SSE 4.1 for a branchless pmind / pmaxd comparator might be interesting, but that would mean always storing and the C source doesn't do that.

If this strategy of double-width load had any merit, it would be better implemented with pure integer on a 64-bit machine like x86-64, where you can operate on just the low 32 bits with garbage (or valuable data) in the upper half. E.g.,

I would use a nested np.where() to eliminate all exact matches first and then ddress the more complex logic you need. I believe this would also be a faster solution as its vectorized than using apply with concat and a custom function. The code would look like this:

Conditional jump instructions do not set flags. So you can for example jump multiple times on the same comparison:

Currently your code will always report BL for greatest! The line mov al,bl ;al keeps the maximum should read mov al, cl.

For AL=max(AL,BL,CL), first find the max between AL and BL and keep it in AL that is to be modified anyway as it will be the result. Then find the max between the possibly new AL and CL yielding the final AL: