0 
for(unsigned int mBlock = 0; mBlock < coords.size(); mBlock++)
{   
    WidgetType widgetType; 
    height = macBlockWidth + coords[mBlock].y;
    width = macBlockHeight + coords[mBlock].x;

    macBlockParent = new QWidget;
    cooefsLink = new QPushButton(macBlockParent);
    macBlock = new QWidget(macBlockParent);
    widgetType.widget = macBlock;
    widgetType.type = (macBlocks[mBlock][2] != 'S') 
                        ? (macBlocks[mBlock][0]) : (macBlocks[mBlock][2]);
    blockWidgetTypes.push_back(widgetType);

    connect(cooefsLink, SIGNAL(released()), 
                                    this, SLOT(showCoefficients()));
    buttonSignals[cooefsLink] = mBlock;

    constructMotionVector(mBlock);
    macBlockLayout->addWidget(macBlockParent, height - 16, width - 16);
    styleMacroBlocks(mBlock);
}

この for ループから関数を作成して、2 つの異なる for ループに分割することで操作を並列化できますか?両方とも同時にベクトルを操作します。1 つは前半の項目に取り組んでおり、2 番目のスレッドは後半を構築しています。たとえば

スレッド 1

for(unsigned int mBlock = 0; mBlock < coords.size() / 2; mBlock++)
{   
    WidgetType widgetType; 
    height = macBlockWidth + coords[mBlock].y;
    width = macBlockHeight + coords[mBlock].x;

    macBlockParent = new QWidget;
    cooefsLink = new QPushButton(macBlockParent);
    macBlock = new QWidget(macBlockParent);
    widgetType.widget = macBlock;
    widgetType.type = (macBlocks[mBlock][2] != 'S') 
                        ? (macBlocks[mBlock][0]) : (macBlocks[mBlock][2]);
    blockWidgetTypes.push_back(widgetType);

    connect(cooefsLink, SIGNAL(released()), 
                                    this, SLOT(showCoefficients()));
    buttonSignals[cooefsLink] = mBlock;

    constructMotionVector(mBlock);
    macBlockLayout->addWidget(macBlockParent, height - 16, width - 16);
    styleMacroBlocks(mBlock);
}

スレッド 2

for(unsigned int mBlock = coords.size() / 2; mBlock < coords.size(); mBlock++)
{   
    WidgetType widgetType; 
    height = macBlockWidth + coords[mBlock].y;
    width = macBlockHeight + coords[mBlock].x;

    macBlockParent = new QWidget;
    cooefsLink = new QPushButton(macBlockParent);
    macBlock = new QWidget(macBlockParent);
    widgetType.widget = macBlock;
    widgetType.type = (macBlocks[mBlock][2] != 'S') 
                        ? (macBlocks[mBlock][0]) : (macBlocks[mBlock][2]);
    blockWidgetTypes.push_back(widgetType);

    connect(cooefsLink, SIGNAL(released()), 
                                    this, SLOT(showCoefficients()));
    buttonSignals[cooefsLink] = mBlock;

    constructMotionVector(mBlock);
    macBlockLayout->addWidget(macBlockParent, height - 16, width - 16);
    styleMacroBlocks(mBlock);
}

それは私のシステムにとって本当のボトルネックであり、1 つの CPU しか使用していないことに気づき、その CPU を使い果たしているためです。どんな助けでも大歓迎です。

4

1 に答える 1

0

うーん...次のような構造がある場合:両方blockWidgetTypes.push_back(widgetType);のスレッドで、マルチスレッドの実行には非常に危険なようです。

于 2013-03-20T21:52:37.487 に答える