二进制“<<”：没有找到接受“Time”类型的右操作数的运算符？

Question

先说明具体问题

    Time aida(3, 35);
    Time temp;
    temp = aida*1.17;
    cout << "Aida * 1.17 = " << temp << endl;
    cout << aida*10.0 << endl;

最后一行的“<<”出现了题目的错误，但是倒数第二行是正常输出的，下面是头文件和源代码

class Time
{
private:
    int hours;
    int minutes;
public:
    Time();
    Time(int h, int min);
    friend Time operator*(Time &a, double n);
    friend ostream & operator<<(ostream &os, Time &a);
};

Time operator*(Time &a, double n)
{
    Time temp;
    long total = a.hours * 60 * n + a.minutes * n;
    temp.hours = total / 60;
    temp.minutes = total % 60;
    return temp;
}
ostream & operator<<(ostream &os, Time &a)
{
    os << a.hours << " hours, " << a.minutes << " minutes.";
    return os;
}

刚开始学不是太懂，希望大家指正一下，谢谢。

Answer 1

#include <iostream>

using namespace std;


class Time
{
private:
    int hours;
    int minutes;
public:
    Time() { hours = 0; minutes = 0; }
    Time(int h, int min) { hours = h; minutes = min; }
    Time& operator*(double n);
    friend ostream & operator<<(ostream &os, Time &a);
};

Time& Time::operator*(double n)
{
    long total = this->hours * 60 * n + this->minutes * n;
    this->hours = total / 60;
    this->minutes = total % 60;
    return *this;
}
ostream & operator<<(ostream &os, Time &a)
{
    os << a.hours << " hours, " << a.minutes << " minutes.";
    return os;
}

int main() {
    Time aida(3, 35);
    Time temp;
    temp = aida*1.17;
    cout << "Aida * 1.17 = " << temp << endl;
    cout << aida*10.0 << endl;
    return 0;
}

Aida * 1.17 = 4 hours, 11 minutes.
41 hours, 50 minutes.

另外，建议你用英文原版的编译器。什么二进制，是Binary，在这里翻译成二元（运算符）。