在下面的代码中，我遇到了名为normalize的函数问题。基本上在这个程序中，我输入一个分数为2个整数。如果输入为负数，则该归一化函数应该使分母为正，并且使分子和/或如果两者都输入为负数，则分母为正。它也应该将分数转换为最低项形式。所以这适用于一些输入而不是其他输入，我不明白为什么。
同样在main函数中，接近结束时我尝试测试这个规范化函数，我不知道如何使用oop原则来做到这一点。我是通过直接引用成员变量来做到的。我怎么能这样做呢？
这是我的课程代码：
class Rational
{
public:
Rational(int numer, int denom);//Initialize an object with 1 arguement for numerator and 1 for denominator
Rational(int wholeNumber);//Initialize an object with 1 argument for numerator. Sets denominator to 1.
Rational();//Initialize an object with 0 for numerator and 1 for denominator
friend ostream& operator <<(ostream& outputStream, const Rational& number);//Overloading of insertion operator
friend istream& operator >>(istream& inputStream, Rational& number);//Overloading of extraction operator
friend bool operator ==(const Rational& number1, const Rational& number2);//Overloading of boolean "=="
friend bool operator <(const Rational& number1, const Rational& number2);//Overloading of inequality sign
friend bool operator <=(const Rational& number1, const Rational& number2);//Overloading of inequality sign
friend bool operator >(const Rational& number1, const Rational& number2);//Overloading of inequality sign
friend bool operator >=(const Rational numerator, const Rational& denominator);//Overloading of inequality sign
friend const Rational operator +(const Rational& number1, const Rational& number2);//Overloading of + operator
friend const Rational operator -(const Rational& number1, const Rational& number2);//Overloading of - operator
friend const Rational operator *(const Rational number1, const Rational& number2);//Overloading of * operator
friend const Rational operator /(const Rational& number1, const Rational& number2);//Overloading of / operator
Rational normalize();//Normalizes the number so fraction is in lowest terms and denominator is positive
int calcGCD(int numerator, int denominator);//Finds greatest common divisor between numerator and denominator
private:
int numerator, denominator;
};
这是规范化功能：
Rational Rational::normalize()
{
if ((numerator > 0 && denominator < 0) || (numerator < 0 && denominator < 0))
{
denominator = -denominator;
numerator = -numerator;
}
int gcd = calcGCD(numerator, denominator);
numerator = numerator/gcd;
denominator = denominator/gcd;
return Rational(numerator, denominator);
}
答案 0 :(得分：1)
以下是一些处理规范化的简单方法。注意除了删除咆哮者之外，我还没有真正检查OP的逻辑。它足以回答OP关于如何在OO环境中正确使用它的问题。
此选项将reference传递给Rational，并创建并返回不同的Rational
Rational normalize(const Rational & in)
{
// get values from Rational to work on so we don't have to keep getting
// them over and over
int num = in.getNumerator();
int den = in.getDenominator();
// numerator and denominator are public, so getters aren't strictly necessary,
// but data members should always be private unless given a damn good reason
// and probably not even then. If you're going to learn, might as well learn
// to do it right, go private and use getters.
if (den < 0)
{ // if the denominator is negative, flip the signs of both den and num
den *= -1;
num *= -1;
}
int gcd = calcgcd(num, den); //grabbing gcd and string in temp
// note I changed the name of the function. int gcd = gcd() is in poor taste even
// if it does compile, and the function should state what it does.
// It doesn't GCD, it calculates GCD.
num = num / gcd;
den = den / gcd;
return Rational(num, den); // create and return a new Rational.
// Compiler should be smart enough to avoid copying
}
使用它：
Rational notNormalized(245, 49);
Rational normalized = normalize(notNormalized);
在这个变体中，Rational使用自己来创建一个不同的，规范化的Rational。
因为这是一个类方法，所以需要将其声明添加到类中。
Rational normalize() const;
const标记表示使用此方法不会以任何方式更改Rational对象。尝试更改normalize方法中的对象将导致编译器错误。
然后执行：
Rational Rational::normalize() const
{
// get copies of numerator and denominator so we don't damage this Rational
int num = numerator;
int den = denominator;
if (den < 0)
{
den *= -1;
num *= -1;
}
int gcd = calcgcd(num, den);
num = num / gcd;
den = den / gcd;
return Rational(num, den);
}
使用它：
Rational notNormalized(245, 49);
Rational normalized = notNormalized.normalize();
在此变体中，Rational规范化自身
因为这是一个类方法，所以需要将其声明添加到类中。
Rational normalize();
然后执行：
void Rational::normalize()
{
if (denominator < 0)
{
denominator *= -1;
numerator *= -1;
}
int gcd = calcgcd(numerator, denominator);
numerator /= gcd;
denominator /= gcd;
}
使用它：
Rational soonToBeNormalized(245, 49);
soonToBeNormalized.normalize();
要快速回答问题，请将问题降至最低限度。类定义，规范化函数和＆＃34;如何使这个规范化函数与这个Rational类一起工作？＆＃34;本来就足够了。其余的是糠，吸引了人们的兴趣，并将帮助人们。