Time addition program

I am a noob programmer. I had a class in regular C programming that didn't prepare me for my C++ programming class. So my knowledge of programming is very limited. With that said, I've got an assginment here that I've started and have no idea how to complete. I was hoping that someone here could make this program. I know what you're thinking: How is that going to help you learn and why should I do it. As to the learning part, I really need to see what this program would look like to learn anything about it. Well, do with this what you'd like:



**************************************************
DESCRIPTION

1.1
Write a program that uses a class.

1.2
Your project will contain files CTime.h, CTime.cpp and TimeApp.cpp.

1.2.1 (Class declaration)

* Place the class declaration and inline functions in a seperate header file, CTime.h.

* Class declaration:

declare CTime as a class
public:

declare a defualt constructor
define an inline destructor with an empty code body
declare getHours as a member function
declare getMinutes as a member function
declare getSeconds as a member function
declare setHours, setMinutes, setSeconds as a member function
declare add as a member function with CTime return and a CTime parameter

private:

declare hours, minutes and seconds as an integer

make the constructor, destructor, get- and set- functions inline.

1.2.2 (Implementation CTime.cpp)

* #include file CTime.h

* place the noninline member function add in a separate file, CTime.cpp

-convert the time values to total seconds before adding them
example:
CTime tm;
long t1Seconds;
t1Seconds = (long)tm.hours*3600 + tm.minutes*60 + tm.seconds;
-add the total seconds together:
t3Seconds = t1Seconds + t2Seconds;
-divide them back out to make the new time hours, minutes and seconds.
- store them in a class CTime variable before returning it.

1.2.3 (member function add)

Module add

Return: CTime

Paramaters: define t2 as a CTime

Data:
define t1Seconds, t2Seconds and t3Seconds as long integers
Processing:

convert hours, minutes and seconds to a long and assign to
t1Seconds

convert t2.hours, t2.minutes and t2.seconds toa long and assign to t2Seconds

SET t3Seconds to t1Seconds + t2Seconds

derive values from t3Seconds to assign to temp.hours, temp.minutes and temp.seconds

return temp

END add

1.2.4 (Application file TimApp.cpp)

this is where the prgoram asks the user for 2 times entered in this format:

12:59:59

the program adds the 2 times and displays the sum


Here's the code I've done:


CTime.h file:

Code:

/******************************* FILE: CTime.h* PROGRAMMER: * DATE: 01/14/09** Declaration of class CTime.*******************************/#ifndef CTime_H#define CTime_H    #include <iostream.h>    class   CTime     {    public:           CTime();           ~CTime()  {}                 int getHours();           int getMinutes();           int getSeconds();           int add();                 void setHours();           void setMinutes();           void setSeconds();                          private:            int hours;            int minutes;            int seconds;    };        inline int CTime::getHours()    {      return hours;         }    inline int CTime::getMinutes()    {      return minutes;        }    inline int CTime::getSeconds()    {      return seconds;        }    inline int CTime::add(CTime)    {      return CTime    }    inline void CTime::setHours(int hr)    {      hours = hr;              }    inline void CTime::setMinutes(int min)    {      minutes = min;             }    inline void CTime::setSeconds(int sec)    {      seconds = sec;             }        #endif

and here's CTime.cpp

Code:

/******************************
* FILE: CTime.cpp
* PROGRAMMER: 
* DATE: 01-14-09
*
* Implementation of class CTime.
*******************************/
#include "CTime.h"
#include <iostream>using namespace std;int main(int argc, char *argv[])
{
    int t1Seconds;
    int t2Seconds;
    int t3Seconds;
    
    int CTime::time()
    {
        t1Seconds = (long)time.hours*3600 + time.minutes*60 + time.seconds;
    
    
                 
    system("PAUSE");
    return EXIT_SUCCESS;
}

 

Doesn't work consistently: 1h 1min + 2h 2min = 3h 3min, but 101+202+40=343 = 3h 43 min.

1h50 + 1h50 + 1h50 = 3h 150 = 5h 30, but 150*3=450; add 40=490. You have to add 40 twice to get the "correct" result of 530.

So for this to work, you have to know how many overflows you've had, and add 40 for each one.

And if you're going to calculate overflows, you may as well convert to minutes, add them up normally, then convert back, e.g. 1h30*3 = 90 min * 3 = 270 min; integer divide by 60=4h; 270-4*60=30 for the minutes.

 

OP: no, you're wrong. You don't learn programming by looking at someone else's completed answer, you learn by doing. That's why the course presents a series of programming exercises for you to complete, rather than just going "Lesson 1, here look at this completed code. Lesson 2, here look at this completed code" and so on. So I'll help you fix the code, but I'm not going to write it for you.

Most of your CTime.cpp is wrong; this file is just for the definition of the CTime class so you don't need any of the io stuff or a main function. I don't know why you've started a CTime::time() function; this is not in the class definition.

What you need to do in CTime.cpp is pretty much already spelled out in detail:

1.2.2 (Implementation CTime.cpp)
* #include file CTime.h
* place the noninline member function add in a separate file, CTime.cpp

So do you know how to declare a non-inline member function? (Hint: it's exactly the same as an inline member function except (a) for the inline keyword and (b) it doesn't go in the header file.)

Your definition in the header of add is slightly wrong:

declare add as a member function with CTime return and a CTime parameter

but you wrote:

inline int CTime::add(CTime) { return CTime }

which has an int return instead of a CTime return, and while it has a CTime parameter that parameter has no name. Also this function (a) is inline, which is not required and (b) has a function body, so when you declare the function in CTime.cpp you will get a duplicate symbol error. The function add() should just be a prototype.

 

Good good......................
You are clever

 

1.50
1.50
1.50
-----
4.50
-----
Now add 0.40

But how do you know that you have to add .40? Dividing by 100 doesn't simplify the problem, which is essentially all you've done.

2.00
1.25
1.25
===
4.50

So do we add .40 or not? Why?

 

Only if what is >=60?
155+155=310, so what do we compare with 60?

 

As per my point of view,...

Example 1h 55m + 1h 50 m
Add 155+155 take 55 + 55 =110 , nCount=0
Now compare it with 60 (if 110 > 60 ) 110 - 60 =50 ++nCoun;
Answer will be (1+1+ ncount)H 50M

Example2:
1h50m + 1h 50m + 1h 50m
=> 50 +50 +50 =150 which is greater than 60 => 150-60 =90 , ncount=1
=> till less than 60 i.e. 90 - 60 =30 , nCount =2
=> So answer will be 1h+1h+1h+2=5h 30m


NOTE: It's slightly diverse from your type of solution ...

 

OK, so what's the point of the 155 value if you're going to reckon hours and minutes separately anyway?

 

Same thing you are doing as adding 1h55m + 1h55m = 310
for i=1 to numberofoverflow=1
add 40 t0 310

result will be : 3h 50m

Example2 : 1h50m + 1h50m +1h50m = 450
for i=1 to numberofoverflow=2
add 40 to 450

result = 5h 30m


Now adding 40 in this case or substracting 60 where you have take minute parametere only .
No difference....

 

How might you implement that in C?

 

Code:

#include<iostream.h>
#include<malloc.h>

#define uint unsigned int
int main()
{
  uint noOfOverFlow=0,reqSum=0,i,hr[3],min[3],minuteSum=0;
   for(i=0;i<3;++i)
   {
     cout<<"Enter "<<i<< " Hour"<<endl;;
     cin>>hr[i];
     cout<<"Enter "<<i<< " minutes"<<endl;
     cin>>min[i];
   }
  minuteSum=min[0]+min[1]+min[2];
  noOfOverFlow = minuteSum/60;
  for(i=0;i<3;++i)
    reqSum= reqSum+ hr[i]*100+min[i];
  cout<<"reqSum"<<reqSum<<endl;
  minuteSum=(reqSum)%100;
    cout<<"MinuteSum"<<minuteSum<<endl;

  for(i=0;i<noOfOverFlow; ++i)
  {
    minuteSum = minuteSum +40;    
  }
 cout<<"MinuteSum"<<minuteSum<<endl;

cout<<"Finally Sum is"<<((hr[0]+hr[1]+hr[2])+ noOfOverFlow)<<"hr"<<minuteSum%100<<endl;;
  
  return 0;
}

Output:

Code:

asadulla ~/temp> g++ test.cc
asadulla ~/temp> ./a.out
Enter 0 Hour
1
Enter 0 minutes
50
Enter 1 Hour
1
Enter 1 minutes
50
Enter 2 Hour
1
Enter 2 minutes
50
reqSum450
MinuteSum50
MinuteSum130
Finally Sum is5hr30
asadulla ~/temp> ./a.out
Enter 0 Hour
1 
Enter 0 minutes
90
Enter 1 Hour
1
Enter 1 minutes
90
Enter 2 Hour
1
Enter 2 minutes
90
reqSum570
MinuteSum70
MinuteSum230
Finally Sum is7hr30

As i will suggest substraccting 60 is better solution than adding 40....
adding 40 method is better if are calculating manually i.e. writing Competetive examination like CAT, GMAT etc.

 

Seems a lot more work than:

Code:

void go4e_15847()
{
  uint noOfOverFlow=0,reqSum=0,i,hr[3],min[3],minuteSum=0;
   for(i=0;i<3;++i)
   {
     cout<<"Enter "<<i<< " Hour"<<endl;;
     cin>>hr[i];
     cout<<"Enter "<<i<< " minutes"<<endl;
     cin>>min[i];
   }
  minuteSum=min[0]+min[1]+min[2]+(hr[0]+hr[1]+hr[2])*60;
  cout<<"Finally sum is "<<(minuteSum/60)<<"h "<<(minuteSum%60)<<"min"<<endl;
}

 

i have already told that ...