成員構造函數定義不正確

Question

我的程序可以運行，但是它告訴我此函數有錯誤

Fraction(int a, int b)
// generate a fraction which is a/b
{
        num = a;
    if(b > 0)
        den = b;
    else
        den = 1;  
}

我的程序是為輸出錯誤而不是由編譯器輸出錯誤而編寫的，因此程序說此成員構造函數不正確，因此我需要確保符號永遠不會為負。 它出什么問題了？

#include <iostream>
#include <cmath>
#include <cassert>

using namespace std;

class Fraction
{
public:
// constructor

Fraction(int a, int b)
// generate a fraction which is a/b
{
        num = a;
    if(b > 0)
        den = b;
    else
        den = 1;  
}

Fraction(int a)
// generate a fraction which is a/1
{
num=a;
den=1;
}

Fraction()
// generate a fraction which is 0/1. i.e 0
{
num=0;
den=1;
}

// member functions

int get_numerator() const
// return the numerator of the fraction
{
return num;
}

int get_denominator() const
// return the denominator of the fraction
{
    return den;
}

void reduce()
// reduce this fraction to simplest form. For instance,
// 2/4 will be reduced to 1/2
{
    num /= gcd();
    den /= gcd();
}

Fraction reciprocal() const
// return the reciprocal of this Fraction
{
    return Fraction(den, num);
}

// friend functions

friend Fraction operator +(const Fraction& f1, const Fraction& f2)
// return the sum of f1 and f2,
// the result is reduced
{
    int n = f1.num * f2.den + f2.num * f1.den;
    int d = f1.den * f2.den;
    Fraction temp = Fraction(n, d);
    temp.reduce();
    return temp;
}

friend Fraction operator -(const Fraction& f1, const Fraction& f2)
// return the difference of f1 and f2,
// the result is reduced
{
    int n = f1.num *f2.den - f2.num * f1.den;
    int d = f1.den * f2.den;
    Fraction temp = Fraction(n, d);
    temp.reduce();
    return temp;
}

friend Fraction operator *(const Fraction& f1, const Fraction& f2)
// return the product of f1 and f2,
// the result is reduced
{
    int n = f1.num * f2.num;
    int d = f1.den * f2.den;
    Fraction temp = Fraction(n, d);
    temp.reduce();
    return temp;
}

friend Fraction operator /(const Fraction& f1, const Fraction& f2)
// return the quotient of f1 and f2,
// the result is reduced
{
    int n = f1.num * f2.den;
    int d = f1.den * f2.num;
    Fraction temp = Fraction(n, d);
    temp.reduce();
    return temp;
}

friend Fraction operator -(const Fraction& f)
// return the negation of f
{
    Fraction temp;
    temp.num=-f.num;
    return temp;
}

friend bool operator < (const Fraction& f1, const Fraction& f2)
// return true if f1 is less than f2.
// False otherwise
{
    return f1.num*f2.den < f2.num*f1.den;
}

friend bool operator > (const Fraction& f1, const Fraction& f2)
// return true if f1 is greater than f2.
// False otherwise
{
    return f1.num*f2.den > f2.num*f1.den;
}

friend bool operator <= (const Fraction& f1, const Fraction& f2)
// return true if f1 is less or equal to f2.
// False otherwise
{
    return f1.num*f2.den <= f2.num*f1.den;
}

friend bool operator >= (const Fraction& f1, const Fraction& f2)
// return true if f1 is greater or equal to f2.
// False otherwise
{
    return f1.num*f2.den >= f2.num*f1.den;
}

friend bool operator == (const Fraction& f1, const Fraction& f2)
// return true if f1 is equal to f2.
// False otherwise
{
    return f1.num*f2.den == f2.num*f1.den;
}

friend bool operator != (const Fraction& f1, const Fraction& f2)
// return true if f1 is not equal to f2.
// False otherwise
{
    return f1.num*f2.den != f2.num*f1.den;
}

friend istream& operator >> (istream& in, Fraction& f)
// input f in the form of a/b, where b cannot be zero. Also,
// if b is negative, the Fraction will change b to be positive.
// So, again, 1/-3 will be changed to -1/3
{
    char temp;
    in >> f.num >> temp >> f.den;
    if(f.den < 0)
    {
        f.num *= -1;
        f.den *= -1;
    }

    return in; 
 }

friend ostream& operator << (ostream& out, Fraction& f)
// output a Fraction f in form of a/b
{
    out << f.num << " / " << f.den;
    return out; 
}


private:
int num; // numerator of the fraction
int den; // denominator of the fraction

int gcd();
// A prvate function that is used to find the gcd of numerator
// and denominator by using Euclidean Algorithm
};

// all following test functions are given

double test1();
// test all constructors and two get methods.
// All these functions worth 1.5 points

double test2();
// test neg, reduce, and reciprocal functions.
// All these functions worth 1.5 points

double test3();
// test add, sub, mul, and div functions.
// All these functions worth 3 points

double test4();
// test less, greater, equal, less_or_equal, greater_or_equal,
// not_equal. All these functions worth 2 points

 double test5();
// test input and output function. This two functions worth 1 points

Answer 1

對於a=3和b=-4您的構造函數將設置num=3和denom=1 。 您需要在調用者端更正期望值或修復代碼以執行以下操作：

if (den > 0){
   num = a;
   den = b;
}
else{
   den = -b;
   num = -a
}

Answer 2

同樣要添加到上面的解決方案中，因此也要處理den == 0的情況。

if(den == 0)
{
  den = 1 ;
}