重载ostream&时出现无效转换错误运算符<<
这是我的 Rational 类的原型
#ifndef RATIONAL_H
#define RATIONAL_H
//forward declaration
class ostream;
class Rational
{
int numerator,denominator;
public:
// the various constructors
Rational();
Rational(int);
Rational(int,int);
//member functions
int get_numerator()const{return numerator;}
int get_denominator()const{return denominator;}
// overloaded operators
// relational operators
bool operator==(const Rational&)const;
bool operator<(const Rational&)const;
bool operator<=(const Rational&)const;
bool operator>(const Rational&)const;
bool operator>=(const Rational&)const;
//arithmetic operators
Rational operator+(const Rational&)const;
Rational operator-(const Rational&)const;
Rational operator*(const Rational&)const;
Rational operator/(const Rational&)const;
//output operator
friend ostream& operator<<(ostream&, const Rational&);
};
#endif //RATIONAL_H
这是重载输出运算符<<的实现在有理数.cpp 中
// friend output operator
ostream& operator<<(ostream& os, const Rational& r)
{
os<<r.get_numerator()<<"/"<<r.get_denominator();
}
，当我尝试编译时，出现以下错误，
g++ -c rational.cpp
rational.cpp: In function ‘ostream& operator<<(ostream&, const Rational&)’:
rational.cpp:81:26: error: invalid conversion from ‘const char*’ to ‘int’ [-fpermissive]
rational.cpp:7:1: error: initializing argument 1 of ‘Rational::Rational(int)’ [-fpermissive]
我希望在将有理数传递给 << 时能够将有理数显示为分子/分母。操作员。
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评论（2）
您的第一个问题是您尝试将 ostream 声明为类。假设您打算使用
std::ostream，您不能这样做，它不合法。其一，它是模板专门化的
typedef，而不是类本身。其次，因为您没有
#include，所以您没有ostream< 的任何标准<<重载的定义/code> 因此，当您尝试<<字符串文字时，编译器会尝试将字符串文字转换为Rational类型，因为这是唯一具有一个<<重载可见。简而言之，您需要
#include并在使用时用std::限定ostream。第三点是您的
operator<<重载需要返回一些内容。您应该附加一个return os;语句，或者简单地return整个流表达式。Your first issue is that you try to forward declare
ostreamas a class. Assuming that you mean to usestd::ostream, you can't do that, its not legal.For one, it's a
typedeffor a template specialization, not a class itself.Second, because you don't
#include <ostream>you don't have a definition for any of the standard<<overloads forostreamso when you try to<<a string literal, the compiler trys to convert the string literal to aRationaltype as that is the only type that has a<<overload visible.Simply, you need to
#include <ostream>and qualifyostreamwithstd::where you use it.A third point is that your overload of
operator<<needs to return something. You should either append areturn os;statement or simplyreturnthe whole streaming expression.虽然我不确定你为什么会收到这样的错误消息，但我确实发现了一个错误：你应该
在你的朋友函数中添加：。
Although I am not sure why you get such error message, I do found an error: you should add:
in your friend function.