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  • Why is ffmpeg faster than this minimal example ?

    23 juillet 2022, par Dave Ceddia

    I'm wanting to read the audio out of a video file as fast as possible, using the libav libraries. It's all working fine, but it seems like it could be faster.

    


    To get a performance baseline, I ran this ffmpeg command and timed it :

    


    time ffmpeg -threads 1 -i file -map 0:a:0 -f null -


    


    On a test file (a 2.5gb 2hr .MOV with pcm_s16be audio) this comes out to about 1.35 seconds on my M1 Macbook Pro.

    


    On the other hand, this minimal C code (based on FFmpeg's "Demuxing and decoding" example) is consistently around 0.3 seconds slower.

    


    #include <libavcodec></libavcodec>avcodec.h>&#xA;#include <libavformat></libavformat>avformat.h>&#xA;&#xA;static int decode_packet(AVCodecContext *dec, const AVPacket *pkt, AVFrame *frame)&#xA;{&#xA;    int ret = 0;&#xA;&#xA;    // submit the packet to the decoder&#xA;    ret = avcodec_send_packet(dec, pkt);&#xA;&#xA;    // get all the available frames from the decoder&#xA;    while (ret >= 0) {&#xA;        ret = avcodec_receive_frame(dec, frame);&#xA;        av_frame_unref(frame);&#xA;    }&#xA;&#xA;    return 0;&#xA;}&#xA;&#xA;int main (int argc, char **argv)&#xA;{&#xA;    int ret = 0;&#xA;    AVFormatContext *fmt_ctx = NULL;&#xA;    AVCodecContext  *dec_ctx = NULL;&#xA;    AVFrame *frame = NULL;&#xA;    AVPacket *pkt = NULL;&#xA;&#xA;    if (argc != 3) {&#xA;        exit(1);&#xA;    }&#xA;&#xA;    int stream_idx = atoi(argv[2]);&#xA;&#xA;    /* open input file, and allocate format context */&#xA;    avformat_open_input(&amp;fmt_ctx, argv[1], NULL, NULL);&#xA;&#xA;    /* get the stream */&#xA;    AVStream *st = fmt_ctx->streams[stream_idx];&#xA;&#xA;    /* find a decoder for the stream */&#xA;    AVCodec *dec = avcodec_find_decoder(st->codecpar->codec_id);&#xA;&#xA;    /* allocate a codec context for the decoder */&#xA;    dec_ctx = avcodec_alloc_context3(dec);&#xA;&#xA;    /* copy codec parameters from input stream to output codec context */&#xA;    avcodec_parameters_to_context(dec_ctx, st->codecpar);&#xA;&#xA;    /* init the decoder */&#xA;    avcodec_open2(dec_ctx, dec, NULL);&#xA;&#xA;    /* allocate frame and packet structs */&#xA;    frame = av_frame_alloc();&#xA;    pkt = av_packet_alloc();&#xA;&#xA;    /* read frames from the specified stream */&#xA;    while (av_read_frame(fmt_ctx, pkt) >= 0) {&#xA;        if (pkt->stream_index == stream_idx)&#xA;            ret = decode_packet(dec_ctx, pkt, frame);&#xA;&#xA;        av_packet_unref(pkt);&#xA;        if (ret &lt; 0)&#xA;            break;&#xA;    }&#xA;&#xA;    /* flush the decoders */&#xA;    decode_packet(dec_ctx, NULL, frame);&#xA;&#xA;    return ret &lt; 0;&#xA;}&#xA;

    &#xA;

    I tried measuring parts of this program to see if it was spending a lot of time in the setup, but it's not – at least 1.5 seconds of the runtime is the loop where it's reading frames.

    &#xA;

    So I took some flamegraph recordings (using cargo-flamegraph) and ran each a few times to make sure the timing was consistent. There's probably some overhead since both were consistently higher than running normally, but they still have the 0.3 second delta.

    &#xA;

    # 1.812 total&#xA;time sudo flamegraph ./minimal file 1&#xA;&#xA;# 1.542 total&#xA;time sudo flamegraph ffmpeg -threads 1 -i file -map 0:a:0 -f null - 2>&amp;1&#xA;

    &#xA;

    Here are the flamegraphs stacked up, scaled so that the faster one is only 85% as wide as the slower one. (click for larger)

    &#xA;

    ffmpeg versus a minimal example, audio from the same file

    &#xA;

    The interesting thing that stands out to me is how long is spent on read in the minimal example vs. ffmpeg :

    &#xA;

    time spent on read call, ffmpeg vs minimal example

    &#xA;

    The time spent on lseek is also a lot longer in the minimal program – it's plainly visible in that flamegraph, but in the ffmpeg flamegraph, lseek is a single pixel wide.

    &#xA;

    What's causing this discrepancy ? Is ffmpeg actually doing less work than I think it is here ? Is the minimal code doing something naive ? Is there some buffering or other I/O optimizations that ffmpeg has enabled ?

    &#xA;

    How can I shave 0.3 seconds off of the minimal example's runtime ?

    &#xA;

  • avcodec/dpx : fix check of minimal data size for unpadded content

    19 octobre 2022, par Jerome Martinez
    avcodec/dpx : fix check of minimal data size for unpadded content
    

    stride value is not relevant with unpadded content and the total count
    of pixels (width x height) must be used instead of the rounding based on
    width only then multiplied by height

    unpadded_10bit value computing is moved sooner in the code in order to
    be able to use it during computing of minimal content size. Also make sure to
    only set it for 10bit.

    Fix 'Overread buffer' error when the content is not lucky enough to have
    (enough) padding bytes at the end for not being rejected by the formula
    based on the stride value

    Fixes ticket #10259.

    Signed-off-by : Jerome Martinez <jerome@mediaarea.net>
    Signed-off-by : Marton Balint <cus@passwd.hu>

    • [DH] libavcodec/dpx.c
  • ffmpeg utility vs direct libav* capabilities

    15 avril 2015, par ransh

    I would like to ask some general question about ffmpeg utility.
    When it is documented in ffmpeg wiki’s :
    https://www.ffmpeg.org/general.html
    That it support some feature (for example jpeg2000), does it mean that only ffmpeg utility support this feature, or does it mean that using directly the libraries in the ffmpeg repository (through libav*, libavcodecand , libopenjpeg, and other external libraries which are part of ffmpeg) will also support this feature as well ?

    Thank you very much,

    Ran