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我正在为我的 C++ 类编写一些代码,但我被困住了。

我实际上正在使用多态动态绑定,我已经成功地将对象存储到我的向量中,现在我想显示存储在向量 ShapeTwoD 中的所有对象。

这是我正在处理的任务的一部分:

Shape2DLin.cpp

void Shape2DLink::Display()
{
    vector<ShapeTwoD*>::iterator vectorIt = shapeobject.begin();
    while(vectorIt != shapeobject.end())
    {
        *vectorIt->view();
        vectorIt++;
    }
}

SquareImp.cpp

void Square::view() const
{
    cout << "Area is: " << area << endl;
}

ShapeTwoDImp.cpp

void Square::view() const
{
    cout << "Area is: " << area << endl;
}

每当我编译应用程序时,我都会收到以下错误。只是想知道为什么会这样?

D:\School\CSCI204-C++\Assignment\Assign 2\Assign2>g++ -o test Shape2Dmain.cpp Sh
ape2DLink.cpp ShapeTwoDImp.cpp SquareImp.cpp RectangleImp.cpp CrossImp.cpp
Shape2DLink.cpp: In member function 'void Shape2DLink::Display()':
Shape2DLink.cpp:94:14: error: request for member 'view' in '* vectorIt.__gnu_cxx
::__normal_iterator<_Iterator, _Container>::operator-> [with _Iterator = ShapeTw
oD**, _Container = std::vector<ShapeTwoD*>, __gnu_cxx::__normal_iterator<_Iterat
or, _Container>::pointer = ShapeTwoD**]()', which is of non-class type 'ShapeTwo
D*'
4

1 回答 1

3

而不是*vectorIt->view()使用(*vectorIt)->view(). 因为operator ->比 dereference/indirection 具有更高的优先级operator *

于 2013-11-06T04:49:11.403 回答