4
\$\begingroup\$

I'm still somewhat new to C, so pardon me for any silly mistakes I've made.

I'm trying to convert video frames into Minecraft map colors using JNI and FFmpeg's av (and swscale) libraries. The code works below but it's sluggish and I'm going to assume it's prone to the branch prediction problem.

The Minecraft map data is stored in ctx->data; it's a 16384 byte array that's fetched from a direct ByteBuffer via GetDirectBufferAddress.

convertRow is called from a thread pool (taken from here); the thread pool has 4 threads to use.

Current specs of the computer:

8GB of RAM

i5 650 @ 3.19GHz

// http://www.codeproject.com/Articles/69941/Best-Square-Root-Method-Algorithm-Function-Precisi
float fast_sqrt(const float x){
    union
    {
        int i;
        float x;
    } u;

    u.x = x;
    u.i = (1<<29) + (u.i >> 1) - (1<<22);
    return u.x;
}

// forgot where this was from; TODO: find source
double dist(unsigned char r, unsigned char g, unsigned char b, Color e2) {
    int rmean = (r + e2.r ) / 2;
    int rDiff = r - e2.r;
    int gDiff = g - e2.g;
    int bDiff = b - e2.b;
    return fast_sqrt((((512+rmean)*rDiff*rDiff)>>8) + 4*gDiff*gDiff + (((767-rmean)*bDiff*bDiff)>>8));
}

char matchColor(unsigned char r, unsigned char g, unsigned char b){
    int index = 0;
    double best = -1;
    double distance;
    for (int i = 4; i < ARR_LENGTH(mapColors); ++i) {
        distance = dist(r, g, b, mapColors[i]);
        if (distance < best || best == -1.0f) {
            best = distance;
            index = i;
        }
    }

    return (char)(index < 143 ? index : -144 + (index - 143));
}

static void convertRow(void* ptr){
    param* params = (param*)ptr;
    int p;
    unsigned char r, g, b;

    for (int y = 0; y < params->ctx->height; y++) {
        p = params->x * 3 + y * params->ctx->rgbFrame->linesize[0];
        r = params->ctx->rgbFrame->data[0][p];
        g = params->ctx->rgbFrame->data[0][p + 1];
        b = params->ctx->rgbFrame->data[0][p + 2];
        params->ctx->data[(params->x / 128) + (params->ctx->width / 128) * (y / 128)][((params->x % 128) + 128 * (y % 128))] = matchColor(r, g, b);
    }

    latch_countdown(params->ctx->latch);
}

JNIEXPORT jboolean JNICALL Java_ga_nurupeaches_vivianmap_natives_NativeVideo_n_1readFrame(JNIEnv* env, jobject jthis, jlong ptr){
    Context* ctx = (Context*)ptr;

    while((ctx->read = av_read_frame(ctx->formatCtx, &(ctx->packetIn))) >= 0){
        if(ctx->packetIn.stream_index == ctx->vtrackId){
            avcodec_decode_video2(ctx->codecCtx, ctx->rawFrame, &(ctx->finished), &(ctx->packetIn));
            if(ctx->finished){
                sws_scale(ctx->sws, (const uint8_t* const*)ctx->rawFrame->data,
                          ctx->rawFrame->linesize, 0, ctx->codecCtx->height,
                          ctx->rgbFrame->data, ctx->rgbFrame->linesize);

                break;
            }
        }
    }

    if(ctx->read < 0){
        // bad. don't do this.
        // stop doing this me.
        Java_ga_nurupeaches_vivianmap_natives_NativeVideo_n_1close(env, jthis, ptr);
    }

    // lock count mutex so that we don't have a race condition
    latch_lockCount(ctx->latch);
    for(int x=0; x < ctx->width; x++){
        param* params = malloc(sizeof(param));
        params->ctx = ctx;
        params->x = x;
        threadpool_add(ctx->pool, convertRow, params, 0);
    }
    // unlock mutex after all tasks have been submitted
    latch_unlockCount(ctx->latch);

    // wait for count to hit 0
    latch_wait(ctx->latch);
    // reset latch
    latch_reset(ctx->latch);
    return true;
}

Any suggestions (especially performance improvements) would be appreciated! I'm willing to edit more code into the question if you need it; I just didn't because I didn't want it to become a needle in a haystack problem.

\$\endgroup\$
3
  • \$\begingroup\$ Welcome to Code Review! Good job on your first answer. \$\endgroup\$
    – SirPython
    Commented Oct 25, 2015 at 22:50
  • 3
    \$\begingroup\$ @SirPython Uh, I think you meant question. But yeah, thanks! \$\endgroup\$
    – Cirno
    Commented Oct 25, 2015 at 22:59
  • \$\begingroup\$ I apologize. I often make this mistake. \$\endgroup\$
    – SirPython
    Commented Oct 26, 2015 at 23:19

1 Answer 1

1
\$\begingroup\$

Iterate on x, not y

This computation loop:

for (int y = 0; y < params->ctx->height; y++) {
    p = params->x * 3 + y * params->ctx->rgbFrame->linesize[0];
    r = params->ctx->rgbFrame->data[0][p];
    g = params->ctx->rgbFrame->data[0][p + 1];
    b = params->ctx->rgbFrame->data[0][p + 2];
    params->ctx->data[(params->x / 128) + (params->ctx->width / 128) *
          (y / 128)][((params->x % 128) + 128 * (y % 128))] = matchColor(r, g, b);
}

is inefficient because you are looping over y whereas your arrays are stored in x order. For example, from the first iteration to the next, you will be reading from:

(y = 0) params->ctx->rgbFrame->data[0][0]
(y = 1) params->ctx->rgbFrame->data[0][linesize]

If you had iterated over x, you would be reading from:

(x = 0) params->ctx->rgbFrame->data[0][0]
(x = 1) params->ctx->rgbFrame->data[0][3]

Similarly, you are writing to:

(y = 0) params->ctx->data[0][0]
(y = 1) params->ctx->data[0][128]

If you iterated over x, you would be writing to:

(x = 0) params->ctx->data[0][0]
(x = 1) params->ctx->data[0][1]

There are two reasons why accessing nearby addresses is faster:

  1. You will be utilizing the CPU cache efficiently by reusing the same cache lines instead of causing lots of cache misses by accessing memory all over the place.
  2. Since your program is multithreaded, your threads will stop contending on the same cache lines. In other words, right now each of your threads is writing to the same cache lines as each other thread, because they are writing to neighboring addresses in the data array. By switching to x order, your threads will each write to their own separate rows in the data array, which means they will be writing to different cache lines from each other (for the most part). This will reduce cache contention and speed things up.
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.