Question

我有一个程序，我正在使用用户指定的测量来计算矩形的面积。我出于特定原因使用类来执行此操作，但我的编译器会生成两个错误...

expression must have class type
left of '.getArea' must have class/struct/union

我该如何解决这个问题？

Rectangle.h

class Rectangle
{
private:
  int length;
  int width;
  int area = length * width;

public:
  Rectangle(int l, int w);
  int getLength();
  void setLength(int l);
  int getWidth();
  void setWidth(int w);
  int getArea();
  void setArea(int a);
};

Rectangle.cpp

Rectangle::Rectangle(int l, int w)
{
   length = l;
   width = w;
}
--some code--
int Rectangle::getArea()
{
   return area;
}
void Rectangle::setArea(int a)
{
   area = a;
}

Area.cpp

int i, lth, wth;

for (i = 0; i < 3; i++)
{
    cout << "Enter your measurements, length first" << endl;
    cin >> lth >> wth;
    Rectangle rMeasure(int lth, int wth);
    cout << "Area of this rectangle is: " << rMeasure.getArea(); //problem code
}

Answer 1

此

Rectangle rMeasure(int lth, int wth);

是一个函数声明。

看来你的意思是

Rectangle rMeasure(lth, wth);

Answer 2

您的代码中很少需要更改

1

private:
    int length;
    int width;
    int area;

2

Rectangle::Rectangle(int l, int w)
{
    length = l;
    width = w;
    this->area=length * width;
}

3

for (i = 0; i < 3; i++)
{
    cout << "Enter your measurements, length first" << endl;
    cin >> lth >> wth;
    Rectangle rMeasure(lth, wth);
    cout << "Area of this rectangle is: " << rMeasure.getArea();
}

Demo

错误：＆＃34;表达式必须具有类类型＆＃34; ＆安培; ＆＃34;左边必须有class / struct / union＆＃34;

2 个答案: