Created
January 17, 2013 16:40
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Fraction class template
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#pragma once | |
#include <stdexcept> | |
// Calculate GCD (for reduction) | |
template<typename T> T gcd(T x, T y) { | |
// Euclidean algorithm | |
T val1 = (x < y) ? y : x, val2 = (x < y) ? x : y; | |
T* m = &val1; T* n = &val2; T* tmpptr = nullptr; | |
while (true) { | |
*m %= *n; | |
if (*m == 0) break; | |
tmpptr = m; m = n; n = tmpptr; | |
} | |
return *n; | |
} | |
// Fraction class template | |
template<typename T> class Rational { | |
private: | |
// Value | |
T num, denom; | |
// Reduction | |
Rational<T>& reduce() { | |
if (denom == 0) throw std::range_error("Denominator is zero"); | |
if (denom < 0) { | |
num *= -1; denom *= -1; | |
} | |
if (num == 0) { | |
denom = 1; | |
} else { | |
T gcd_ans = gcd<T>(num, denom); | |
num /= gcd_ans; denom /= gcd_ans; | |
} | |
return *this; | |
} | |
public: | |
// Constructors | |
Rational(T numerator = 0, T denominator = 1) { | |
if (denominator == 0) throw std::invalid_argument("Cannot set denominator to 0"); | |
num = numerator; denom = denominator; | |
reduce(); | |
} | |
Rational(const Rational<T>& arg) { | |
num = arg.num; denom = arg.denom; | |
} | |
Rational<T>& operator=(const Rational<T>& arg) { | |
num = arg.num; denom = arg.denom; | |
return *this; | |
} | |
Rational<T>& operator=(T arg) { | |
num = arg; denom = 1; | |
return *this; | |
} | |
// Unary plus and minus | |
const Rational& operator+() const {return *this;} | |
const Rational& operator-() const { | |
return Rational<T>(-num, denom).reduce(); | |
} | |
// addition | |
const Rational& operator+(const Rational<T>& addend) const { | |
return Rational<T>(num * addend.denom + addend.num * denom, denom * addend.denom).reduce(); | |
} | |
Rational& operator+=(const Rational<T>& addend) { | |
return (*this = Rational<T>(num * addend.denom + addend.num * denom, denom * addend.denom).reduce()); | |
} | |
// Subtraction | |
const Rational& operator-(const Rational<T>& subtrahend) const { | |
return Rational<T>(num * subtrahend.denom - subtrahend.num * denom, denom * subtrahend.denom).reduce(); | |
} | |
Rational& operator-=(const Rational<T>& subtrahend) { | |
return (*this = Rational<T>(num * subtrahend.denom - subtrahend.num * denom, denom * subtrahend.denom).reduce()); | |
} | |
// Multiplication | |
const Rational& operator*(const Rational<T>& multiplier) const { | |
return Rational<T>(num * multiplier.num, denom * multiplier.denom).reduce(); | |
} | |
Rational& operator*=(const Rational<T>& multiplier) { | |
return (*this = Rational<T>(num * multiplier.num, denom * multiplier.denom).reduce()); | |
} | |
// Division | |
const Rational& operator/(const Rational<T>& divisor) const { | |
if (multiplier.num == 0) throw std::invalid_argument("Division by zero"); | |
return Rational<T>(num * divisor.denom, denom * divisor.num).reduce(); | |
} | |
Rational& operator/=(const Rational<T>& divisor) { | |
if (multiplier.num == 0) throw std::invalid_argument("Division by zero"); | |
return (*this = Rational<T>(num * divisor.denom, denom * divisor.num).reduce()); | |
} | |
// Equality | |
bool operator==(const Rational<T>& arg) const { | |
return (num == arg.num) && (denom == arg.denom); | |
} | |
bool operator!=(const Rational<T>& arg) const {return !(*this == arg);} | |
// Comparison | |
bool operator<(const Rational<T>& arg) const { | |
return ((num * arg.denom) < (arg.num * denom)); | |
} | |
bool operator>=(const Rational<T>& arg) const {return !(*this < arg);} | |
bool operator>(const Rational<T>& arg) const {return (arg < *this);} | |
bool operator<=(const Rational<T>& arg) const {return !(arg < *this);} | |
// Accessor | |
T getNumerator() const {return num;} | |
T getDenominator() const {return denom;} | |
// Cast into float | |
operator float() const {return (float)num / (float)denom;} | |
operator double() const {return (double)num / (double)denom;} | |
operator long double() const {return (long double)num / (long double)denom;} | |
}; |
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I really like your template. Do you happen to know why the multiplier.num leads to an error in the if statement for division. Thank you so much for your help.