Question

Edit: 增加更多信息。

我愿轮流使用<条码>QPolygonF。 a. 来源如下:

“blocktations”/

象这个多功能(图像中心是来源——<代码>0, 0):

https://i.stack.imgur.com/08ImV.png” alt=“first polygon”/>

To rotate clockwise so that it ends up here:

“entergraph

www.un.org/Depts/DGACM/index_french.htm 上面图象中第一组位置相同:

QPolygonF p1 =  QPolygonF() << QPointF(0, 1) << QPointF(4, 1) << QPointF(4, 2) << QPointF(0, 2);

接着,我轮值了我认为正确的来源(2, 2):

QTransform t;
t.translate(2, 2);
t.rotate(-90);
t.translate(-2, -2);
QPolygonF p2 = t.map(p1);
qDebug() << p1 << "rotated = " << p2;

产出:

QPolygonF(QPointF(0, 1) QPointF(4, 1) QPointF(4, 2) QPointF(0, 2) ) rotated =  QPolygonF(QPointF(1, 4) QPointF(1, 0) QPointF(2, 0) QPointF(2, 4) )

当我想要的产出是:

QPolygonF(QPointF(0, 1) QPointF(4, 1) QPointF(4, 2) QPointF(0, 2) ) rotated =  QPolygonF(QPointF(2, 0) QPointF(3, 0) QPointF(3, 4) QPointF(2, 4) )

但是,根据上述产出,多角将最终考虑:

“wrong的轮调”/

在什么时候,我应轮流?

Answer 1

<><>Edit>>

在阐述你之后,我知道实现这一点的正确守则:

QTransform t; 
t.translate(2, 2); 
t.rotate(90); 
t.translate(-2, -2); 
QPolygonF p2 = t.map(p1); 
qDebug() << p1 << "rotated = " << p2;

这将使问题更加棘手:

QPolygonF(QPointF(3, 0) QPointF(3, 4) QPointF(2, 4) QPointF(2, 0) )

这正是你所期望的(尽管由于轮换而改变次序)。值得注意的是,<代码>rotate(90)和rotate(-90)与你轮流使用直径,而不是middle。

接受QT tranforms

Some confusion may arise because of the direction of the qt rotate function. Although this will enact a counter clockwise rotation, because we "see" the y-axis inverted (at least in this case) it appears clockwise to us.

As an analogy if I m in a room and observer a lamp falling over to the left, from the point of view of someone hanging on the cieling it has "fallen over" to the right.

Answer 2

我只是简短的眼光,但我认为这是错误的:

t.translate(2, 2);

To obtain your transformation should be:

t.translate(2, 1);

从我的理解看, rec子的下限为1。

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