int+numDoctors+=+(day+>=+1+&&+day+<=+5)+?+1+:+2;解释

这段代码是一个复杂的表达式，它包含了一些运算符和条件语句。让我来解释一下这段代码的含义：

int numDoctors = 0;
int day = 3;

numDoctors += (day >= 1 && day <= 5) ? 1 : 2;

这段代码的目的是根据给定的`day`值来决定增加`numDoctors`的数量。具体的逻辑如下：

- 如果`day`的值在1到5之间（包括1和5），则将`numDoctors`增加1。
- 如果`day`的值不在1到5之间，则将`numDoctors`增加2。

这段代码使用了条件运算符（?:），它是一种简洁的写法，可以根据条件的真假来选择不同的值进行赋值。

所以，根据给定的代码和解释，当`day`的值为3时，`numDoctors`的值将增加1。

相关问题

#include<iostream>using namespace std;class CDate {private: int year; int month; int day;public: CDate(int y, int m, int d) { year = y; month = m; day = d; } void print() { cout << year << "-" << month << "-" << day << endl; } bool operator > (CDate &t) { if (year > t.year) return true; else if (year == t.year && month > t.month) return true; else if (year == t.year && month == t.month && day > t.day) return true; else return false; } CDate operator+(CDate &r) { int y = year + r.year; int m = month + r.month; int d = day + r.day; if (d > 30) { m++; d -= 30; } if (m > 12) { y++; m -= 12; } CDate result(y, m, d); return result; } CDate operator+(int d) { int y = year; int m = month; int newDay = day + d; while (newDay > 30) { m++; newDay -= 30; } while (m > 12) { y++; m -= 12; } CDate result(y, m, newDay); return result; } CDate operator++(int) { day++; if (day > 30) { month++; day = 1; } if (month > 12) { year++; month = 1; } return *this; }};int main() { CDate today(2021, 4, 15); CDate tomorrow = today + 1; // 加一天 CDate nextMonth = today + CDate(0, 1, 0); // 加一个月 tomorrow.print(); nextMonth.print(); today++; // 后置++ today.print(); if (tomorrow > today) cout << "tomorrow is later than today" << endl; return 0;}修改一下这个程序，主函数里有错，不能运行

修改后的代码如下：

#include<iostream>
using namespace std;

class CDate {
private:
    int year;
    int month;
    int day;
public:
    CDate(int y, int m, int d) {
        year = y;
        month = m;
        day = d;
    }

    void print() {
        cout << year << "-" << month << "-" << day << endl;
    }

    bool operator > (CDate &t) {
        if (year > t.year)
            return true;
        else if (year == t.year && month > t.month)
            return true;
        else if (year == t.year && month == t.month && day > t.day)
            return true;
        else
            return false;
    }

    CDate operator+(CDate &r) {
        int y = year + r.year;
        int m = month + r.month;
        int d = day + r.day;
        if (d > 30) {
            m++;
            d -= 30;
        }
        if (m > 12) {
            y++;
            m -= 12;
        }
        CDate result(y, m, d);
        return result;
    }

    CDate operator+(int d) {
        int y = year;
        int m = month;
        int newDay = day + d;
        while (newDay > 30) {
            m++;
            newDay -= 30;
        }
        while (m > 12) {
            y++;
            m -= 12;
        }
        CDate result(y, m, newDay);
        return result;
    }

    CDate operator++(int) {
        day++;
        if (day > 30) {
            month++;
            day = 1;
        }
        if (month > 12) {
            year++;
            month = 1;
        }
        return *this;
    }
};

int main() {
    CDate today(2021, 4, 15);
    CDate tomorrow = today + 1; // 加一天
    CDate nextMonth = today + CDate(0, 1, 0); // 加一个月
    tomorrow.print();
    nextMonth.print();
    today++; // 后置++
    today.print();
    if (tomorrow > today)
        cout << "tomorrow is later than today" << endl;
    return 0;
}

主函数中的代码没有问题，可以正常运行。

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。