运营商==超载功能失败

Question

我超载运营商><==在一个简单的时间等级。

double exchange_time::seconds_from_midnight() const { 
    return seconds + minutes * 60.0 + hours * 3600.0 + milliseconds/1000.0; 
} 

bool exchange_time::operator<(const exchange_time& other) const 
{ 
    return seconds_from_midnight() < other.seconds_from_midnight(); 
} 

bool exchange_time::operator>(const exchange_time& other) const 
{ 
    return seconds_from_midnight() > other.seconds_from_midnight(); 
} 

bool exchange_time::operator==(const exchange_time& other) const 
{ 
    return seconds_from_midnight() == other.seconds_from_midnight(); 
} 

该>和<工作完美。然而==产生错误，我的测试失败：

TEST_F(exchange_time_test, comparison) { 
    exchange_time et1, et2; 
    et1.set("93500001"); 
    et2.set("93500123"); 
    EXPECT_TRUE(et2 > et1); 
    EXPECT_TRUE(et1 < et2); 
    EXPECT_TRUE(et2 == et2); 
} 

有什么我失踪了吗？

这里是我的声明：

class exchange_time { 
    public: 
     void set(string timestamp); 
     unsigned short int get_milliseconds() { return milliseconds; } 
     unsigned short int get_seconds() { return seconds; } 
     unsigned short int get_minutes() { return minutes; } 
     unsigned short int get_hours() { return hours; } 
     double seconds_from_midnight() const; 
     bool operator<(const exchange_time& other) const; 
     bool operator>(const exchange_time& other) const; 
     bool operator==(const exchange_time& other) const; 
    private: 
     unsigned short int milliseconds; 
     unsigned short int seconds; 
     unsigned short int minutes; 
     unsigned short int hours; 
}; 

Answer 1

决不比较为双号的平等。检查它们是否几乎相等。最常见的方法是使用epsilon来比较值。

bool exchange_time::operator==(exchange_time other) 
{ 
    return abs(seconds_from_midnight() - other.seconds_from_midnight()) < EPS; 
} 

凡EPS是一个很小的值。如果你需要准确的比较，你需要定义你自己的分数类。

EDIT
EPS代表ε，其被定义为通过的Dictionary.com东西一个非常小的，不显着，或可忽略的量。