Question

好的，这是我编辑的代码，但现在每次编译时都会出现Segmentation Fault（核心转储）。我哪里出错了？我不确定我是否正确地“回溯”了它，但这是我从尝试中得到的结果：编程接收信号SIGSEGV，分段错误。 MyTime :: MyTime中的0x000109f0（this =， h =，m =）在MyTime.cc：10 10 MyTime :: MyTime（int h，int m）{

//MyTime.h File
#include <iostream>

class MyTime
{
  public:

   MyTime(int h = 0, int m = 0);

   void Reset(int h, int m);

   void input();

   void output() const;

   MyTime operator + (const MyTime& t1) const;

   MyTime operator - (const MyTime& t1) const;

   MyTime operator * (const int& num) const;

   MyTime operator / (const int& num) const;

   bool operator == (const MyTime& t1) const;

   bool operator < (const MyTime& t1) const;

   bool operator <= (const MyTime& t1) const;

   int get_hours() const{return hours;}
   int get_minutes() const{return minutes;}

 private:
    void simplify();
    int hours;      // hours can be > 24
    int minutes;   // 0 <= minutes <= 59
};

    std::istream& operator >>(std::istream& fin, MyTime& t);

    std::ostream& operator <<(std::ostream& fout, const MyTime& t);

//MyTime.cc File
#include "MyTime.h"
#include <cstdlib>
#include <iostream>
#include <iomanip>
using namespace std;

// Constructor

MyTime::MyTime(int h, int m){
    hours = h;
    minutes = m;
}

void MyTime::Reset(int h, int m){
    hours = h;
    minutes = m;
}

void MyTime::simplify(){
    hours += minutes/60;
    minutes = minutes%60;
}

void MyTime::input(){
    char junk;
    cin >> hours;
    cin.get(junk);
    cin >> minutes;
    simplify();
}

void MyTime::output() const{
    cout << hours << ':' << setw(2) << setfill('0') << minutes;
}

MyTime MyTime::operator +(const MyTime& t1) const{
    MyTime tmp;
    tmp.hours = t1.hours + hours;
    tmp.minutes = t1.minutes + minutes;
    tmp.simplify();
    return tmp;
}

MyTime MyTime::operator -(const MyTime& t1) const{
    MyTime tmp;
    tmp.minutes = abs((t1.hours*60+t1.minutes) -
                                    (hours*60+minutes));
    tmp.simplify();
    return tmp;
}

MyTime MyTime::operator /(const int& num) const{
    MyTime tmp;
    tmp.minutes = hours * 60 + minutes;
    tmp.minutes /= num;
    tmp.simplify();
    return tmp;
}

MyTime MyTime::operator *(const int& num) const{
    MyTime tmp;
    tmp.minutes = hours * 60 + minutes;
    tmp.minutes *= num;
    tmp.simplify();
    return tmp;
}

bool MyTime::operator == (const MyTime& t1) const{
    return t1.hours == hours && t1.minutes == minutes;
}

bool MyTime::operator < (const MyTime& t1) const{
    return (t1.hours * 60 + t1.minutes) < (hours * 60 + minutes);
}

bool MyTime::operator <=(const MyTime& t1) const{
    return (t1 == (hours * 60 + minutes)) || (t1 < (hours * 60 + minutes));
}

ostream& operator <<(ostream& fout, const MyTime& t){
    t.output();
    return fout;
}

istream& operator >>(istream& fin, MyTime& t){
    t.input();
    return fin;
}

//main.cc File
#include <iostream>
#include "MyTime.h"

int main()
{
   MyTime t1, t2;
   int scalar;

   std::cout << "Enter a time:  ";
   std::cin >> t1;

   std::cout << "Enter another time:  ";
   std::cin >> t2;

   std::cout << "Enter a scalar to manipulate those times:  ";
   std::cin >> scalar;

   if(t1 == t2)
     std::cout << t1 << " is equal to " << t2 << std::endl;

   if(t1 < t2)
     std::cout << t1 << " is less than " << t2 << std::endl;

   if(t1 <= t2)
     std::cout << t1 << " is less than or equal to " << t2 << std::endl;

     std::cout << t1 << " + " << scalar << " = " << t1 + scalar << std::endl;
     std::cout << t1 << " - " << scalar << " = " << t1 - scalar << std::endl;
     std::cout << t1 << " * " << scalar << " = " << t1 * scalar << std::endl;
     std::cout << t1 << " / " << scalar << " = " << t1 / scalar << std::endl;

     std::cout << t2 << " + " << scalar << " = " << t2 + scalar << std::endl;
     std::cout << t2 << " - " << scalar << " = " << t2 - scalar << std::endl;
     std::cout << t2 << " * " << scalar << " = " << t2 * scalar << std::endl;
     std::cout << t2 << " / " << scalar << " = " << t2 / scalar << std::endl;

     return 0;
}

Answer 1

首先，整理单位，例如：时间乘以或除以时间是没有意义的。添加或减去时间的时间给出了时间。时间乘以或除以数字给出时间。

接下来，在内部将时间存储为分钟。转换为人类消费的小时和分钟。

之后其余部分将落实到位。

Answer 2

对于乘法和除法，如果时间完全正确，则MyTime :: simplify不执行任何操作。如果0 <=分钟＆lt;分钟/ 60将总是为零。 60，因为分钟和60都是整数，所以它会向下舍入。类似地，只要相同的条件成立，分钟％60将等于分钟。除此之外，我同意这样的评论：一次又一次地划分是......奇怪的，你应该避免的事情。

对于第二个问题，请切换

(t1.hours * 60 + t1.minutes) < (hours * 60 + minutes)

到

(hours * 60 + minutes) < (t1.hours * 60 + t1.minutes)

以便运营商＆lt;对

this < t1

实际上在操作员的左侧对此进行评估，因为您已将其置于过载状态。

如何乘以和除以一天的时间，例如10:00和5:00有超载功能？

2 个答案: