Question

我想迭代一个Object列表列表，

list<list<Point>> gnuPoints;

这是一个静态列表，它接受来自圆和多边形等各种类的列表，所有类都推送多个对象，即Point（x，y）到列表

 while (angle != 360)
{
 double x = m_radius*cos(angle);
 double y = m_radius*sin(angle);
 angle += 30;
 circlePoint.push_back(Point(x, y));
}

similarly goes with Polygon and line shapes

polygonPoint.push_back(Point(x,y));
linePoint.Push_back(point(x,y));

然后将这些列表推送到gnuPoint（list＆lt; list＆lt; Point＆gt;＆gt;）。

gnuPoints.push_back(circlePoints);
gnuPoints.push_back(polygonPoints);
gnuPoints.push_back(linePoints);

现在我想在文件中写出不同形状的所有x，y值，为了迭代它，我无法在此代码之后找到任何特定的解决方案。

for (list<list<Point>>::iterator it = Point::gnuPoints.begin();
    it != Point::gnuPoints.end();
    it++)
{
//My assumption is that another For loop would come but could not apply 
  as I don't know what is available at the first index of gnuPoints list.
}

Answer 1

for (list<list<Point>>::iterator it = Point::gnuPoints.begin();
    it != Point::gnuPoints.end();
    it++)
{
     for (list<Point>::iterator innerIt = it->begin(); innerIt != it->end(); ++innerIt)
}

如果您已经使用支持c ++ 11的编译器，则可以简单地将其写为：

for (const auto& container : Point::gnuPoints)
{
    for (const auto& item : container)
    {
    }
}

迭代对象列表

1 个答案: