用户选择选项后如何使用switch语句进行函数调用？

Question

我试图在这里进行函数调用，但是我不知道出了什么问题。我收到一个错误：

使用未初始化的局部变量'区域'

我们必须使用switch大小写。如果有人可以帮助解决此问题，那就太好了。

// Include Section
#include <iostream>
#include <cstdlib>
using namespace std;

//Function prototype
double rectangle(double, double);
double triangle(double, double);
double circle(double);

// Main Program
int main()
{
    //Variable declaration
    int choice;
    const double PI = 3.14159; //pi
    double area;
    double length, //length for rectangle
            width; //width for rectangle
    double height; //height for triangle
    double base; //base for triangle
    double radius; //radius for circle

    //Give choices
    cout << "Welcome to Geometry Calculator \n";
    cout << "Pick one option from the following: \n";
    cout << "1. Calculate the Area of a Rectangle \n";
    cout << "2. Calculate the Area of a Triangle \n";
    cout << "3. Calculate the Area of a Circle \n";
    cout << "4. Quit \n\n";

    //input choice
    cout << "Enter your choice (1-4): ";
    cin >> choice;

    //Input Validation
    while (choice <= 0 || choice > 4)
    {
        cout << "Please enter a valid menu chice: ";
        cin >> choice;
    }

    //Switch statement
    switch (choice)
    {
        case 1: //rectangle
            cout << "Enter the LENGTH of your Rectangle: ";
            cin >> length;
            cout << "Enter the WIDTH of your Rectangle: ";
            cin >> width;
            rectangle(length, width);
            //function
            break;

        case 2: //Triangle
            cout << "Enter the LENGTH of the base of your Triangle: ";
            cin >> base;
            cout << "Enter the HEIGHT of your Triangle: ";
            cin >> height;
            triangle(base, height);
            return 0;
            break;

        case 3: //Circle
            cout << "Enter the RADIUS of your Circle: ";
            cin >> radius;
            circle(radius);
            return 0;
            //calculation
            break;

        case 4: //quit
            cout << "You chose to quit. Thanks for using my program! \n\n";
    }
    return 0;
}

//function call
void triangle(double base, double height);
{
    area = base * height * 0.5;
    cout << "The AREA of your Triangle is " << area << ". \n\n";
}

void circle(double radius);
{
    area = PI * radius * radius;
    cout << "The AREA of your Circle is " << area << ". \n\n";
}

double fc = rectangle(double length, double width);
{
    area = length * width;
    cout << "The area of the rectangle is " << area << endl;
}

Answer 1

您在这里有多个问题：

---

在函数定义上，尾随的分号不应存在：

void triangle(double base, double height);  // <-- remove this semicolon
{
    area = base * height * 0.5;
    cout << "The AREA of your Triangle is " << area << ". \n\n";
}

此错误在其他定义上也存在。分号终止函数前向声明（在文件顶部），但是在定义函数时，括号将其替换。

---

您还声明了三个函数triangle /circle/ rectangle以返回double，但是在实现中，您将它们定义为返回不匹配的void。这可能是您收到的特定错误的来源。

更改前向声明以匹配定义，反之亦然。

---

这三个函数还试图分配一个不在范围内的area变量。如果要在其中使用此变量，则必须在这些函数中声明它。

请注意，这些功能无法访问area中的main()变量。

---

rectangle的定义以变量声明开头，这没有任何意义：

double fc = rectangle(double length, double width);
这被解析为全局fc变量的声明，该声明已初始化为调用rectangle函数的结果，但是参数语法不正确，因此解析将在那里失败。大概应该这样写：

void rectangle(double length, double width)
或

double rectangle(double length, double width)
功能circle()引用符号PI，但这不在范围内。 （您在PI内部声明了main()，因此它仅在该函数中可见。您可以通过将其移至全局范围来解决此问题。）