为什么我没有编码时我的班级正在执行算术？

Question

我正在做一个家庭作业项目，旨在向我们介绍课程。我刚刚开始，但由于某种原因，我的成员变量一直正确显示到最后。它应该只是在函数之间传递这些变量然后输出它们。它将采用参数化构造函数设置的分子和分母，用它们设置一个对象，然后输出分子和分母所持有的内容。它正在获取分子并将其设置为零，即使它之前在该类运行的每个函数中都显示了正确的数字。分母更奇怪，因为它以某种方式变成了一个巨大的数字。我只是想了解发生了什么，因为我是新手，以及为什么对我来说不是很明显。

#include <iostream>

using std::cout;
using std::endl;
using std::ostream;

class RationalNum
{
public:
    RationalNum();
    RationalNum(int numer, int denom);
    int getNumerator() const { return numer; }
    int getDenominator() const { return denom; }
    void setNumerator(int);
    void setDenominator(int);
    void set(int& numer, int& denom);
    void output(ostream& outs);

private:
    int denom;
    int numer;
};

RationalNum::RationalNum() : numer(1), denom(1)
{}

RationalNum::RationalNum(int numer, int denom) //takes two integers for numer and denom, and reduces
{
    cout << "The numer and denom  in the param const are " << numer << denom << endl;
    set(numer, denom);
}

void RationalNum::set(int& numer, int& denom) //use other functions to take two params and set as numer and denom

{
    cout << "The numer and denom in set are " << numer << denom << endl;
    setNumerator(numer);
    setDenominator(denom);
}

void RationalNum::setNumerator(int numer) //takes an int, sets it as numer, and reduces it
{
    cout << "in setNumer, the numer is: " << numer << endl;
    //reduce(newNumer); //Not using yet
}

void RationalNum::setDenominator(int denom) //takes an int, sets it as denom, and reduces it
{
    cout << "in setDenom, the denom is: " << denom << endl;
    //reduce(newDenom); //Not using yet
}

void RationalNum::output(ostream& outs) //takes an ostream param
{
    cout << numer << "/" << denom << endl;
}


int main()
{
    RationalNum rn1, rn2(25, 10), rn3(24, -100);

    cout << "rn1 contains: ";
    rn1.output(cout);
    cout << endl;

    cout << "rn2 contains: ";
    rn2.output(cout);
    cout << endl;

    cout << "rn3 contains: ";
    rn3.output(cout);
}

这是实际结果。直到最后两行都是正确的，这就是我迷路的地方。 rn2 应显示 25/10，rn3 应显示 24/-100

The numer and denom  in the param const are 2510
The numer and denom in set are 2510
in setNumer, the numer is: 25
in setDenom, the denom is: 10

The numer and denom  in the param const are 24-100
The numer and denom in set are 24-100
in setNumer, the numer is: 24
in setDenom, the denom is: -100
rn1 contains: 1/1    
rn2 contains: 0/4196160    
rn3 contains: 0/4197376

Answer 1

setNumerator 和 setDenominator 方法没有做任何事情。由于这个原因，类成员denom 和numer 并没有被初始化。因此，它们的值将是未定义的，即可能是垃圾。

要修复它，请使用：

void RationalNum::setNumerator(int n)
{
  numer = n; // I changed the parameter name to avoid confusion with the member
  cout << "in setNumer, the numer is: " << numer << endl;
  //reduce(newNumer); //Not using yet
}

对分母使用类似的方法。