Question

嗨我收到了我尝试过的“函数未在此范围内声明”的每个函数的错误。我是C ++的新手，并试图让我的TA解决这个问题，但仍然没有想到它。谢谢你的帮助！

这是我的头文件（circle.h）：

#ifndef _CIRCLE_H_
#define _CIRCLE_H_

class circle
{
private:
    double x1;
    double x2;
    double y1;
    double y2;
protected:
    double distance(double x1, double y1, double x2, double y2);
public:
    double radius(double x1, double y1, double x2, double y2);
    double circumference(double x1, double y1, double x2, double y2);
    double area(double x1, double y1, double x2, double y2);
    void populate_classobj(double x1, double y1, double x2, double y2);
};
#endif

这是我定义类函数的地方（circle.cc）：

#include <cmath>
#include "circle.h"
#define PI (4*atan(1))
using namespace std;

double circle::distance(double x1, double y1, double x2, double y2) {
    double dist;
    dist = sqrt(pow((x2-x1),2)+pow((y2-y1),2));
    return dist;
}

double circle::radius(double x1, double y1, double x2, double y2) {
    return distance(x1, y1, x2, y2);
}

double circle::area(double x1, double y1, double x2, double y2) {
    double circle_area = PI*(pow(radius(x1, y1, x2, y2),2));
    return circle_area;
}

double circle::circumference (double x1, double y1, double x2, double y2) {
    double circ = 2*PI*radius(x1, y1, x2, y2);
    return circ;
}

void circle::populate_classobj (double x_1, double y_1, double x_2, double y_2) {
    double x1 = x_1;
    double y1 = y_1;
    double x2 = x_2;
    double y2 = y_2;
}

这是我的主要（main.cc）：

#include <iostream>
#include <cmath>
#include "circle.h"
using namespace std;

int main()
{
    double x_1, y_1, x_2, y_2;
    int switch_val;
    cout << "Enter x1 coordinate: ";
    cin >> x_1;
    cout << "Enter y1 coordinate: ";
    cin >> y_1;
    cout << "Enter x2 coordinate: ";
    cin >> x_2;
    cout << "Enter y2 coordinate: ";
    cin >> y_2;
    circle mycircle;
    mycircle.populate_classobj (x_1, y_1, x_2, y_2);
    do{
        cout << "1 for radius, 2 for circumference, 3 for area, 4 for exit" << endl;
        cout << "Enter your desired computation: ";
        cin >> switch_val;
        switch(1) {
            case 1: cout << "Radius is: " <<endl;
                    cout << radius(x_1, y_1, x_2, y_2);
                    break;
            case 2: cout << "Circumference is: " <<endl;
                    cout << circumference (x_1, y_1, x_2, y_2);
                    break;
            case 3: cout << "Area is: " <<endl;
                    cout << area(x_1, y_1, x_2, y_2);
                    break;
            case 4: cout << "Exiting program... "<< endl;
                    break;
            default: cout << "Invalid input please re-input value"<<endl;
                    break;
        }
    }while(switch_val != 4);
    return 0;
}

我认为这与我的#include陈述有关，但我不确定。谢谢！

Answer 1

在switch中，您确实使用了课程circle的方法，例如radius。类的功能（所谓的方法）仅可用于使用该类的现有实例进行调用。

E.g。

circle mycircle;
double radius = mycircle.radius(1.0,2.0,3.0,4.0);

会奏效。

但是如果方法不需要该类的任何成员变量，则可以将其声明为static。只需在标题中的函数前写下单词：

`static double radius(double x1, double y1, double x2, double y2);`

现在您可以在没有circle的实例的情况下调用它。

double radius = circle::radius(1.0,2.0,3.0,4.0);

您只需使用classes命名空间，让编译器知道函数的来源。

C ++链接标题/主要错误

1 个答案: