根据QPixmap 类参考:
QPixmap 对象可以通过值传递,因为 QPixmap 类使用隐式数据共享。有关详细信息,请参阅隐式数据共享文档。
QPixmap 实现:
QPixmap::QPixmap(const QPixmap &pixmap)
: QPaintDevice()
{
if (!qt_pixmap_thread_test()) {
init(0, 0, QPixmapData::PixmapType);
return;
}
if (pixmap.paintingActive()) { // make a deep copy
operator=(pixmap.copy());
} else {
data = pixmap.data;
}
}
只有当像素图处于活动状态时,您才需要深拷贝,否则新像素图只需要复制原始数据指针。
对于 const 引用和指针的区别:
QPixmap largeMap(1000, 1000);
QPainter p(&largeMap);
int count = 100000;
qint64 time1, time2;
QPixmap *pSmallMap = new QPixmap("e:/test.png");
QPixmap smallMap = QPixmap("e:/test.png");
time1 = QDateTime::currentMSecsSinceEpoch();
for (int i = 0; i < count; ++i) {
p.drawPixmap(0, 0, *pSmallMap);
}
time2 = QDateTime::currentMSecsSinceEpoch();;
qDebug("def time = %d\n", time2 - time1);
time1 = QDateTime::currentMSecsSinceEpoch();
for (int i = 0; i < count; ++i) {
p.drawPixmap(0, 0, smallMap);
}
time2 = QDateTime::currentMSecsSinceEpoch();;
qDebug("normal time = %d\n", time2 - time1);
在 Visual Studio 2010调试配置下编译将产生以下程序集:
28: p.drawPixmap(0, 0, *pSmallMap);
003B1647 8B 55 C4 mov edx,dword ptr [ebp-3Ch] //the pixmap pointer
003B164A 52 push edx
003B164B 6A 00 push 0 //x
003B164D 6A 00 push 0 //y
003B164F 8D 4D F0 lea ecx,[ebp-10h] //the qpainter pointer
003B1652 FF 15 9C D7 3B 00 call dword ptr [__imp_QPainter::drawPixmap (3BD79Ch)]
35: p.drawPixmap(0, 0, smallMap);
003B16A8 8D 4D E0 lea ecx,[ebp-20h] //the pixmap pointer
003B16AB 51 push ecx
003B16AC 6A 00 push 0 //x
003B16AE 6A 00 push 0 //y
003B16B0 8D 4D F0 lea ecx,[ebp-10h] //the qpainter pointer
003B16B3 FF 15 9C D7 3B 00 call dword ptr [__imp_QPainter::drawPixmap (3BD79Ch)]
这两者之间应该没有区别,因为编译器将生成相同的汇编代码:将指针传递给 drawPixmap 函数。
QDateTime::currentMSecsSinceEpoch() 在我的盒子上几乎显示了相同的结果。