CODEWITHSUNDEEP
c++performancemicrobenchmark
rgkirch
rgkirch

Reputation: 35

in c++, the fastest way to convert a positive number to 1 and negative to 0

I'm writing this code to make a candlestick chart and I want a red box if the open price for the day is greater than the close. I also want the box to be green if the close is higher than the open price.

if(open > close) {
    boxColor = red;
} else {
    boxColor = green;
}

Pseudo code is easier than an English sentence for this.

So I wrote this code first and then tried to benchmark it but I don't know how to get meaningful results.

for(int i = 0; i < history.get().close.size(); i++) {
    auto open = history->open[i];
    auto close = history->close[i];
    int red = ((int)close - (int)open) >> ((int)sizeof(close) * 8);
    int green = ((int)open - (int)close) >> ((int)sizeof(close) * 8);
    gl::color(red,green,0);
    gl::drawSolidRect( Rectf(vec2(i - 1, open), vec2(i + 1, close)) );
}

This is how I tried to benchmark it. Each run just shows 2ns. My main question to the community is this:

Can I actually make it faster by using a right shift and avoid a conditional branch?

#include <benchmark/reporter.h>

static void BM_red_noWork(benchmark::State& state) {
    double open = (double)rand() / RAND_MAX;
    double close = (double)rand() / RAND_MAX;
    while (state.KeepRunning()) {
    }
}
BENCHMARK(BM_red_noWork);

static void BM_red_fast_work(benchmark::State& state) {
    double open = (double)rand() / RAND_MAX;
    double close = (double)rand() / RAND_MAX;
    while (state.KeepRunning()) {
        int red = ((int)open - (int)close) >> sizeof(int) - 1;
    }
}
BENCHMARK(BM_red_fast_work);

static void BM_red_slow_work(benchmark::State& state) {
    double open = (double)rand() / RAND_MAX;
    double close = (double)rand() / RAND_MAX;
    while (state.KeepRunning()) {
        int red = open > close ? 0 : 1;
    }
}
BENCHMARK(BM_red_slow_work);

Thanks!

Upvotes: 1

Views: 1318

Answers (1)

OmnipotentEntity
OmnipotentEntity

Reputation: 17131

As I stated in my comment, the compiler will do these optimizations for you. Here is a minimal compilable example:

int main() {
  volatile int a = 42;
  if (a <= 0) {
    return 0;
  } else {
    return 1;
  }
}

The volatile is simply to prevent optimizations from "knowing" the value of a and instead it forces it to be read.

This was compiled with the command g++ -O3 -S test.cpp and it produces a file named test.s

Inside test.s is the assembly generated by the compiler (pardon AT&T syntax):

movl    $42, -4(%rsp)
movl    -4(%rsp), %eax
testl   %eax, %eax
setg    %al
movzbl  %al, %eax
ret

As you can see, it is branchless. It uses testl to set a flag if the number is <= 0 and then reads that value using setg, moves it back into the proper register, then finally it returns.

It should be noted, at this was adapted from your code. A much better way to write this is simply:

int main() {
  volatile int a = 42;
  return a > 0;
}

It also generates the same assembly.

This is likely to be better than anything readable you could write directly in C++. For instance your code (hopefully corrected for bit arithmetic errors):

int main() {
  volatile int a = 42;
  return ~(a >> (sizeof(int) * CHAR_BIT - 1)) & 1;
}

Compiles to:

movl    $42, -4(%rsp)
movl    -4(%rsp), %eax
notl    %eax
shrl    $31, %eax
ret

Which is indeed, very slightly smaller. But it's not significantly faster. Especially not when you have a GL call right next to it. I'd rather spend 1-3 additional cycles to get readable code, rather than have to scratch my head wondering what my coworker (or me from 6 months ago, which is essentially the same thing) did.

EDIT: I should be remarked that the compiler also optimized the bit arithmetic I wrote, because I wrote it less well than I could have. The assembly is actually: (~a) >> 31 which is equivalent to the ~(a >> 31) & 1 that I wrote (at least in most implementations with an unsigned integer, see comments for details).

Upvotes: 4

Related Questions