完成日期类CDate{ CDate(); CDate (int year,int month,int day); CDate operator ++(); CDate operator ++(int); CDate operator +(long days) const; CDate operator -(long days) const; long operator -(const CDate &rhs) const; int WeekDay() const; static

&dt) const;}的定义中，operator-()返回两个日期之间相差的天数，您能否给出具体实现方法？

可以的。具体的实现方法如下：

#include<cmath>
class CDate{
    int year,month,day;
public:
    CDate();
    CDate(int year,int month,int day);
    CDate operator+(int days);
    CDate operator-(int days);
    int operator-(CDate otherDate);
};

CDate::CDate(){
    year=2000;
    month=1;
    day=1;
}

CDate::CDate(int year,int month,int day){
    this->year=year;
    this->month=month;
    this->day=day;
}

CDate CDate::operator+(int days){
    CDate result=(*this);
    int i;
    for(i=1;i<=days;i++){
        result.day++;
        if(result.day>days_in_month[result.month]+(result.month==2&&isleapyear(result.year))){
            result.day=1;
            result.month++;
            if(result.month>12){
                result.month=1;   
                result.year++;
            }
        }
    } 
    return result;
}

CDate CDate::operator-(int days){
    CDate result=(*this);
    int i;
    for(i=1;i<=days;i++){
        result.day--;
        if(result.day<1){
            result.month--;
            if(result.month<1){
                result.month=12;
                result.year--;
            }
            result.day=days_in_month[result.month]+(result.month==2&&isleapyear(result.year));
        }
    } 
    return result;
}

int CDate::operator-(CDate otherDate){
    int i,days=0;
    if(otherDate>(*this)){
        for(i=year;i<otherDate.year;i++){
            days+=isleapyear(i)?366:365;
        }
        for(i=1;i<otherDate.month;i++){
            days-=days_in_month[i]+(otherDate.year%4==0&&otherDate.month==2);
        }
        days-=otherDate.day;
        for(i=1;i<month;i++){
            days+=days_in_month[i]+(year%4==0&&month==2);
        }
        days+=day;
    }else{
        for(i=otherDate.year;i<year;i++){
            days+=isleapyear(i)?366:365;
        }
        for(i=1;i<month;i++){
            days-=days_in_month[i]+(year%4==0&&month==2);
        }
        days-=day;
        for(i=1;i<otherDate.month;i++){
            days+=days_in_month[i]+(otherDate.year%4==0&&otherDate.month==2);
        }
        days+=otherDate.day; 
    }
    return days;
}

其中，`days_in_month`数组记录了每个月的天数，`isleapyear()`函数判断某一年是否是闰年。实现中，我们先判断目标日期与当前日期之间相差多少天，然后再根据目标日期是前还是后，分别计算即可。