第九周任务二 (这是个问题)
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#include <iostream> using namespace std; class CTime { private: unsigned short int hour; // 时 unsigned short int minute; // 分 unsigned short int second; // 秒 public: CTime(int h=0,int m=0,int s=0):hour(h), minute(m), second(s){ }; void setTime(int h,int m,int s); void display(); //比较运算符(二目)的重载 bool operator > (CTime &t); bool operator < (CTime &t); bool operator >= (CTime &t); bool operator <= (CTime &t); bool operator == (CTime &t); bool operator != (CTime &t); //二目运算符的重载 CTime operator+(CTime &c);//返回c所规定的时、分、秒后的时间,例t1(8,20,25),t2(11,20,50),t1+t2为:41:15 CTime operator-(CTime &c);//对照+理解 CTime operator+(int s);//返回s秒后的时间 CTime operator-(int s);//返回s秒前的时间 //一目运算符的重载 CTime operator++(int);//后置++,下一秒 CTime operator++();//前置++,下一秒 CTime operator--(int);//后置--,前一秒 CTime operator--();//前置--,前一秒 //赋值运算符的重载 CTime operator+=(CTime &c); CTime operator-=(CTime &c); CTime operator+=(int s);//返回s秒后的时间 CTime operator-=(int s);//返回s秒前的时间friend ostream & operator <<(ostream &,CTime &); friend istream & operator >>(istream &,CTime &);}; //下面实现所有的运算符重载代码。 //为简化编程,请注意通过调用已有函数,利用好各函数之间的关系 void CTime::setTime(int h,int m,int s) { hour=h; minute=m; second=s; } void CTime::display() { cout<<hour<<":"<<minute<<":"<<second<<endl; } bool CTime::operator > (CTime &t) { if(hour>t.hour) return true; else if(minute>t.minute) return true; else if(second>t.second) return true; else return false; } bool CTime::operator < (CTime &t) { if(hour<t.hour) return true; else if(minute<t.minute) return true; else if(second<t.second) return true; else return false; } bool CTime::operator >= (CTime &t) { if(operator <(t)!=1) return true; else return false; } bool CTime::operator <= (CTime &t) { if(operator >(t)!=1) return true; else return false; } bool CTime::operator == (CTime &t) { if(operator <(t)!=1||operator >(t)!=1) return true; else return false; } bool CTime::operator != (CTime &t) { if(operator <(t)==1||operator >(t)==1) return true; else return false; } CTime CTime::operator+(CTime &c) { CTime c1; c1.second=second+c.second; c1.minute=minute+c.minute; c1.hour=hour+c.hour; while(c1.second>59) { c1.second=c1.second-60; c1.minute=c1.minute+1; } while(c1.minute>59) { c1.minute=c1.minute-60; c1.hour=c1.hour+1; } while(c1.hour >= 24)c1.hour = c1.hour - 24; return c1; } CTime CTime::operator-(CTime &c) { CTime c1; int h = 0,m = 0,s;s = c.second + c.minute* 60 + c.hour *3600 - second - minute * 60 - hour * 3600;if(s < 0)s = -s;while(s > 59){s = s - 60;m ++;}while(m > 59){m = m - 60;h ++;}while(h >= 24)h = h - 24;c1.hour = h;c1.minute = m;c1.second = s;return c1; } CTime CTime::operator+(int s) { CTime c1; c1.second = second + s; c1.hour = hour;c1.minute = minute; while(c1.second>59) { c1.second = c1.second - 60; ++c1.minute; } while(c1.minute>59) { c1.minute = c1.minute - 60; ++c1.hour; } while(c1.hour >= 24)c1.hour = c1.hour; return c1; } CTime CTime::operator-(int s) { CTime c1; c1.second = second;c1.minute = minute;c1.hour = hour; while(c1.second<59) { c1.second+=60; --c1.minute; } while(c1.minute<0) { c1.minute+=60; --c1.hour; } c1.second=second-s; return c1; } CTime CTime::operator++(int) { CTime temp(*this); second++; while(second>59) { second-=60; ++minute; } return temp; } CTime CTime::operator++() { while(++second>59) { second-=60; ++minute; } return *this; } CTime CTime::operator--(int) { CTime temp(*this); second--; return temp; } CTime CTime::operator--() { --second; return *this; } CTime CTime::operator+=(CTime &c) { CTime c1; c1.second = second + c.second; c1.minute = minute + c.minute; c1.hour = hour + c.hour; while(c1.second > 59) { c1.second = c1.second - 60; c1.minute ++; } while(c1.minute>59) { c1.minute=c1.minute-60; c1.hour ++; } while(c1.hour >= 24)c1.hour = c1.hour - 24;*this = c1; //问题所在点 return c1; } CTime CTime::operator-=(CTime &c) { CTime c1; int h = 0,m = 0,s;s = c.second + c.minute* 60 + c.hour *3600 - second - minute * 60 - hour * 3600;if(s < 0)s = -s;while(s > 59){s = s - 60;m ++;}while(m > 59){m = m - 60;h ++;}while(h >= 24)h = h - 24;c1.hour = h;c1.minute = m;c1.second = s;*this = c1;}CTime CTime::operator+=(int s) { CTime c1; c1.second = second + s; c1.hour = hour;c1.minute = minute; while(c1.second>59) { c1.second = c1.second - 60; ++c1.minute; } while(c1.minute>59) { c1.minute = c1.minute - 60; ++c1.hour; } while(c1.hour >= 24)c1.hour = c1.hour;*this = c1; return c1; } CTime CTime::operator-=(int s) { CTime c1; c1.second = second;c1.minute = minute;c1.hour = hour; if(c1.second<59) { c1.second+=60; --c1.minute; } if(c1.minute<0) { c1.minute+=60; --c1.hour; } c1.second=second-s; *this = c1; return c1; } ostream & operator <<(ostream & output,CTime & c){output<<"("<<c.hour<<":"<<c.minute<<":"<<c.second<<endl;return output;} istream & operator >>(istream &input,CTime & c) { char d; cout<<"请输入时间(hh:mm:ss)"; input>>c.hour>>d>>c.minute>>d>>c.second; return input; }void main() { CTime t1(8,20,25), t2(11,20,50), t; cout << "t1为:"<<t1; cout << "t2为:"<<t2; cout << "比较两个时间大小:" << endl; if (t1 > t2) cout << "t1 > t2" << endl; if (t1 < t2) cout << "t1 < t2" <<endl; if (t1 == t2) cout <<"t1 = t2" <<endl; if (t1 != t2) cout << "t1 ≠ t2" << endl; if (t1 >= t2) cout << "t1 ≥ t2" << endl; if (t1 <= t2) cout << "t1 ≤ t2" << endl; cout<<endl; t = t1 + t2; cout << "t1 + t2 = "<<t; t = t1-t2; cout << "t1-t2 = "<<t; t = t1; t = t + 35; cout<< "t1 + 35 = "<<t; t = t2; t = t - 8; cout<< "t2 - 8 = "<<t ; t = t1; t = t ++; cout << endl; cout << "t1++ = "<<t; t = t1; t = ++t; cout << "++t1 = "<<t; t = t2; t = t --; cout << "t2-- = "<<t; t = t2; t = --t; cout << "--t1 = "<<t; t = t1; t += t2; cout << "t1 += t2,t1 = " << t; t = t2; t -= t1; cout << "t2 -= t1,t2 = "<<t ; t = t1; t += 5; cout << "t1 += 5 = "<<t ; t = t1; t -= 6; cout << "t1 -= 6 = " <<t; system("pause"); }问题所在:这是麻烦方法实现的任务二,本应该是上周提出的问题
请看代码:
CTime CTime::operator+=(CTime &c) { CTime c1; c1.second = second + c.second; c1.minute = minute + c.minute; c1.hour = hour + c.hour; while(c1.second > 59) { c1.second = c1.second - 60; c1.minute ++; } while(c1.minute>59) { c1.minute=c1.minute-60; c1.hour ++; } while(c1.hour >= 24)c1.hour = c1.hour - 24;*this = c1; return c1; }注意倒数第三行(*this = c1;),若没有这一行,则主函数中输出的时间是没经过改变的,仍然是t1的原值,有了这行代码就对了。很多函数斗神这样,想了很长时间没有明白,还望明天老师能够指点迷津。
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