Question

任务是在上面的代码中编写计算新点的成员函数，这是另外两个点的数量。而且我不知道如何返回物体或我应该怎么做。这是代码，功能标有三个!!!该函数必须返回一些东西，我不能使它无效，因为不允许引用void。

<form id="basic" method="post" name="basic"> 
<select class="select-style gender" name="eid">
        <option value="blank">Please Select E-MITRA ID</option>
        <?php
 while($row=mysqli_fetch_array($result))
 {
?>
  <option value="<?php echo $row['uid']; ?>"><?php echo $row['uid']." (" . $row['name'] .")"; ?></option>
<?php
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?>
</select>

    <p class="contact"><label for="bid">Micro-ATM Serial No</label></p> 
<input type="text" name="miatm" value ="<?php if (isset($miatm)) echo $miatm; ?>"    /> <?php echo $miErr; ?>

 <p class="contact"><label for="bid">Micro-ATM TID No</label></p> 
<input type="text" name="tid" value ="<?php if (isset($tid)) echo $tid; ?>" /> <?php echo $tiErr; ?>

   <input class="buttom" name="submit" id="submit"  value="Add Me" type="submit">

我做到了！这是必须添加的功能

    class point {
    private:
        float x;
        float y;
    public:
        point();
        point(float xcoord, float ycoord);
        void print();
        float dist(point p1, point p2);
    !!! float &add(point p1, point p2);
        float  &X();
        float &Y();
        ~point();
    };
    float & point::X() { return x; }
    float & point::Y() { return y; }
    point::point() {
        cout << "Creating POINT (0,0)" << endl;
        x = y = 0.0;
    }
    point::point(float xcoord, float ycoord) {
        cout << "Creating POINt (" << xcoord << "," << ycoord << ")" << endl;
        x = xcoord;
        y = ycoord;
    }
    void point::print() {
        cout << "POINT (" << x << "," << y << ")";
    }
    float point::dist(point p1, point p2) {
        return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y));
    }
   !!!// float & point::add(point p1, point p2) {
        point z;
        z.X() = p1.X() + p2.X();
        z.Y() = p1.Y() + p2.Y();
        z.print();
    }
    point::~point() {
        cout << "Deleting ";
        print();
        cout << endl;
    }
    int main() {
        point a(3, 4), b(10, 4);
        cout << "Distance between"; a.print();
        cout << " and "; b.print();
        cout << " is " << a.dist(a, b) << endl;
    }

非常感谢ForceBru !!和你们所有人

Answer 1

怎么做

您也可以返回point：

point point::add(const point& p1, const point& p2) {
    point result;

    result.x = p1.x + p2.x;
    result.y = p1.y + p2.y;

    return result;
}

请注意，这里不需要使用X()和Y()函数，因为此方法已经可以访问私有成员。

也可以进行操作员重载

/*friend*/ point operator+ (const point& one, const point& two) {
    // the code is the same
}

如何使用

int main() {
    point one(2,5), two(3,6), three;

    three.add(one, two);

    std::cout << "Added " << one.print() << " and " << two.print();
    std::cout << " and got " << three.print() << std::endl;

    return 0;
}

修改：正如评论中所述，您不应该返回对add函数中创建的对象的引用，因为在这样的情况下您只允许返回对类成员和static变量的引用的情况。

Answer 2

您可以在此处使用运算符重载：

point point::operator+(const point & obj)  {

point obj3;
obj3.x = this->x + obj.x;
return obj3;

}

添加两点后返回对象。

简单示例：

class Addition  {
int a;
public:
      void SetValue(int x);
      int GetValue();
      Addition operator+(const Addition & obj1);
};
void Addition::SetValue(int x)
{
    a = x;
}
int Addition::GetValue()
{
   return a;
 }
Addition Addition::operator+(const Addition &obj1)
{
  Addition obj3;
  obj3.a = this->a + obj1.a;
  return obj3;
 }

int _tmain(int argc, _TCHAR* argv[])
{
  Addition obj1;
  int Temp;
  std::cout<<"Enter Value for First Object : "<<std::endl;
  std::cin>>Temp;

  obj1.SetValue(Temp);

  Addition obj2;
  std::cout<<"Enter Value for Second Object : "<<std::endl;
  std::cin>>Temp;

  obj2.SetValue(Temp);

  Addition obj3;
  obj3 = obj1 + obj2;
  std::cout<<"Addition of point is "<<obj3.GetValue()<<std::endl;

  return 0;

}

从C ++中的类成员函数返回对象

2 个答案:

怎么做

如何使用