Question

我想问一下问题在哪里？

我收到了错误：'unary *'的无效类型参数

我是c ++编程的新手，我使用java风格。指针和解引用对我来说是个大问题。

我的应用程序得到输入值并保存为点对象，此后我应该找到2行的交集。

我想要返回一个Point对象，在其中我将计算x和y值。

.h文件

class Point {
public:
    double x_val, y_val;

    Point(double, double);

    double x();
    double y();

    double dist(Point other);

    Point add(Point b);
    Point sub(Point b);

    void move(double a, double b);



};



 class Triungle {
public:


        Triungle(std::string);
        void compute_length();
        void lines_intersect(Point a, Point b, Point c, Point d, Point *intersection);

        Point a, b, c;
};

.cpp文件

Point::Point(double x = 0.0, double y = 0.0) {
    x_val = x;
    y_val = y;
}

double Point::x() {
    return x_val;
}

double Point::y() {
    return y_val;
}

double Point::dist(Point other) {
    double xd = this->x() - other.x();
    double yd = this->y() - other.y();
    return sqrt(xd * xd + yd * yd);
}

Point Point::add(Point b) {
    return Point(x_val + b.x_val, y_val + b.y_val);
}

Point Point::sub(Point b) {
    return Point(x_val - b.x_val, y_val - b.y_val);
}

void Point::move(double a, double b) {
    x_val += a;
    y_val += b;
}

    void Triungle::lines_intersect(Point a, Point b, Point c, Point d, Point *intersection) {

        double x, y;
        double A1 = b.y_val - a.y_val;
        double B1 = b.x_val - a.x_val;
        double C1 = a.y_val - (A1 / B1) * a.x_val;

        double A2 = d.y_val - c.y_val;
        double B2 = d.x_val - c.x_val;
        double C2 = c.y_val - (A2 / B2) * c.x_val;

        double det = (A1 / B1) - (A2 / B2);
        if (det == 0) {
            // lines are paralel
        } else {
            x = (C2 - C1) / det;
            y = (A1 * C2 - A2 * C1) / det;
        }
        *intersection->x_val = x;  // here i got error
        *intersection->y_val = y;  // here i got error
       }

Triungle::Triungle(std::string s) {

    cout << "enter first point of " << s << " triangle: ";
    cin >> a.x_val;
    cin >> a.y_val;
    if (!(cin)) {
        cout << "input error." << endl;
        exit(1);
    }
    cin.clear();
    cout << "enter second point of " << s << " triangle: ";
    cin >> b.x_val;
    cin >> b.y_val;
    if (!(cin)) {
        cout << "input error." << endl;
        exit(1);
    }
    cin.clear();
    cout << "enter 3 point of " << s << " triangle: ";
    cin >> c.x_val;
    cin >> c.y_val;
    if (!cin) {
        cout << "input error." << endl;
        exit(1);
    }
    cin.clear();

}

我以这种方式调用函数

 int main(int argc, char** argv) {
    Triungle a("first");
    Triungle b("second");
 Point p;

    a.lines_intersect(a.a, a.b, a.c, a.a, &p);

}

Answer 1

intersection->member

将取消引用指针intersection。这与

相同

(*intersection).member

你不需要取消引用它两次。

Answer 2

您在代码中执行的操作

*intersection->x_val = x;

等同于

(*(intersection->x_val)) = x;

因为通过指针->选择的运算符比解除引用运算符*具有更高的precedence，后者的优先级高于赋值运算符=。

首先，您选择double x_val班的成员Point。其次，您尝试将一元解除引用运算符*应用于结果。并且因为x_val是double，而不是指针，这是取消引用运算符所期望的，编译器报告错误。

因此，取消引用运算符在这里过多，并且足以执行以下操作

intersection->x_val = x;

Answer 3

假设你得到的错误是两行的编译错误：

*intersection->x_val = x;  // here i got error 
*intersection->y_val = y;  // here i got error

问题是您要取消引用指针，然后在其上使用derefencing运算符->。

相反，你应该这样做：

intersection->x_val = x; 
intersection->y_val = y;  // leave it as a pointer

或

*intersection.x_val = x; 
*intersection.y_val = y;  // use it as an object

C ++段线交叉

3 个答案: