尽管我提供了参数，为什么我在这里得到垃圾值？

Question

#include <iostream>
#include <math.h>
#include <conio.h>
using namespace std;

class Triangle
{
public:
    int a, b, c;
    void getdata();
    int peri(int a, int b, int c)
    {
        return a + b + c;
    }
    float s;
    float area;
    float Area(int a, int b, int c)
    {
        s = (a + b + c) / 2;
        area = sqrt(s * (s - a) * (s - b) * (s - c));
        cout << area;
    }

};

void Triangle::getdata()
{
    cin >> a >> b >> c;
}

int main()
{
    int x, y, z;
    cout << "Enter the three sides";
    Triangle t1;
    t1.getdata();
    cout << "The area of that triangle is " << t1.Area(x, y, z) << "and the perimeter "
         << t1.peri(x, y, z);
    return 0;
}

实际上，尽管没有编译错误，但这段代码给了我垃圾值。它没有给我所需的面积和周长输出。那么为什么我会得到垃圾值呢？

Answer 1

这里从三角形初始化a、b、c ：

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void Triangle::getdata( )\n{\n   cin>>a>>b>>c;\n}\n

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但是这里你使用了没有初始化的x,y,z ，就有问题了

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int main()\n{\n   int x,y,z; //<-- these three variables are nowhere initialised\n   cout<<"Enter the three sides";\n   Triangle t1;\n   t1.getdata();\n   cout<<"The area of that triangle is "\n      <<t1.Area( x, y, z) //where x, y, z are initialised??\n      <<"and the perimeter " \n      <<t1.peri(x,y,z);  //where x, y, z are initialised?\n   return 0;\n}\n

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更新 1：正如 @Bathsheba 所指出的， s = (a + b + c) / 2;将由于整数除法而截断。哎呀。使用 0.5f * (a + b + c)。\xe2\x80\x93

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这就是你应该如何实现你的三角形

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class Triangle\n{\npublic:\n    int a, b, c;\n    void getdata();\n    int peri() //remove the arguments, the class has its own members to use\n    {\n        return a + b + c;\n    }\n    float s;\n    float area;\n    float Area() //remove the arguments, the class has its own members to use\n    {\n        s = 0.5f * (a + b + c); //As pointed by @Bathsheba\n        area = sqrt(s * (s - a) * (s - b) * (s - c));\n        cout << area;\n    }\n\n};\n

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然后这样写main函数

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int main()\n{\n   //no additional x, y, z\n   cout<<"Enter the three sides";\n   Triangle t1;\n   t1.getdata();\n   cout<<"The area of that triangle is "\n      <<t1.Area() //no arguments\n      <<"and the perimeter " \n      <<t1.peri();  //no arguments\n   return 0;\n}\n

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