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  #1  
Old 12-Nov-2008, 08:41
minwei86 minwei86 is offline
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Add days to date without affecting the original input date

I know how to add days to date, but i just wan to display the after-added date. I still need the original date. What can i go about it

CPP / C++ / C Code:
void Date::addDay()
{
    int DaysInMonth[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    string months [12] = {"jan", "feb", "mar", "apr", "may", "jun", "jul", "aug", "sep", "oct", "nov", "dec"};


    for(int i=0;i<12;i++)
    {
    if(mth == months[i])
    count = i;
    }
    bool flag;
    int theDays;
    cout<<"Enter how many days to be added"<<endl;
    cin>>theDays;"\n";

    if(isLeapYear())
    DaysInMonth[1] = 29;
    int tempDays = day;
    tempDays += theDays;

    do
    {
        flag = false;
        if(tempDays>DaysInMonth[count])
        {
            tempDays-=DaysInMonth[count];
            count++;
            flag = true;
        }
        if(count > 12)
        {
            count=1;
            year++;
        }
    }
    while(flag);
    day = tempDays;

    cout<<year<<"/"<<day<<"/";
    switch(count)
    {
        case 0 : cout<<"jan"<<endl;
                 break;
        case 1 : cout<<"feb"<<endl;
                 break;
        case 2 : cout<<"mar"<<endl;
                 break;
        case 3 : cout<<"apr"<<endl;
                 break;
        case 4 : cout<<"may"<<endl;
                 break;
        case 5 : cout<<"jun"<<endl;
                 break;
        case 6 : cout<<"jul"<<endl;
                 break;
        case 7 : cout<<"aug"<<endl;
                 break;
        case 8 : cout<<"sep"<<endl;
                 break;
        case 9 : cout<<"oct"<<endl;
                 break;
        case 10 : cout<<"nov"<<endl;
                 break;
        case 11 : cout<<"dec"<<endl;
                 break;
    }


}


void Date::returnDate() const
{
    cout<<"date is "<<day<<"-"<<mth<<"-"<<year<<endl;
}

When i input a date, it will do all sort of operations before going to returnDate() to print out.

I able to calculate the add days to date, but i dun wan it to affect the input date when i print out. What should i do?
  #2  
Old 12-Nov-2008, 10:05
minwei86 minwei86 is offline
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Re: add days to date without affecting the original input date

anyone have some idea or hint to give me?

when i enter example 20 jan 2008...then i add 20 days to it...the result is 9 feb 2008...but i wan when i prompt to print the date, it still contain 20 jan 2008
  #3  
Old 12-Nov-2008, 14:03
minwei86 minwei86 is offline
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Re: Add days to date without affecting the original input date

anyone can help...give me some advice also can...
  #4  
Old 12-Nov-2008, 15:34
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TurboPT TurboPT is offline
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Re: Add days to date without affecting the original input date

One possible way is to store the original day into a temporary variable, do the necessary calculations using that variable instead. This will leave the original value intact.

Oh, and just for suggestion -- instead of repeating the month's in a hard-coded way, because you have declared an array of months, consider replacing the switch with something like this:
CPP / C++ / C Code:
if ( count >= 0 && count < 12 ) // verify the index range!
{
     cout << months[count] << endl;
}
else
{
     cout << "Err, uh, the month is out of range?" << endl;
}
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  #5  
Old 12-Nov-2008, 15:55
ocicat ocicat is offline
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Re: add days to date without affecting the original input date

Quote:
Originally Posted by minwei86
when i enter example 20 jan 2008...then i add 20 days to it...the result is 9 feb 2008...but i wan when i prompt to print the date, it still contain 20 jan 2008
As you are beginning to find out, dealing with calendar dates isn't necessarily trivial.

Personally, I think you will find it simpler to exploit the functions found in ctime.h -- especially the functions:Examples can be found at the links given above.
  #6  
Old 12-Nov-2008, 20:49
minwei86 minwei86 is offline
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Re: Add days to date without affecting the original input date

Quote:
Originally Posted by TurboPT
One possible way is to store the original day into a temporary variable, do the necessary calculations using that variable instead. This will leave the original value intact.

Oh, and just for suggestion -- instead of repeating the month's in a hard-coded way, because you have declared an array of months, consider replacing the switch with something like this:
CPP / C++ / C Code:
if ( count >= 0 && count < 12 ) // verify the index range!
{
     cout << months[count] << endl;
}
else
{
     cout << "Err, uh, the month is out of range?" << endl;
}


how to store it inside a temp variable can teach me, demo me example.
  #7  
Old 13-Nov-2008, 04:33
minwei86 minwei86 is offline
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Re: Add days to date without affecting the original input date

anyone know how to store original day into temp variable...can someone guide me
  #8  
Old 13-Nov-2008, 09:43
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TurboPT TurboPT is offline
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Re: Add days to date without affecting the original input date

Again, this is only one possibility...
CPP / C++ / C Code:
int day = 0;  // your name might differ, but assume this exists somewhere?

// somewhere appropriate...(another variable for a 'temp', call it whatever you want)
int another_day = 0;

// then merely use 'day' in the expression...
// do whatever calc. is needed...
// THIS IS ONLY A BASIC EXAMPLE ( considering your 20 days, from Jan 20, or 'months[0]' )
another_day = (day + 20) % months[0]; // 9, assuming day=20

// TO DO, for you: handling the proper month/year to go along with the 'another_day' change.
Maybe that'll give you a start?

EDIT:
Like ocicat already pointed out, this should starting 'painting the picture' of date handling not being so trivial.
Unless you are limited by the assignment, check into the functions that he offered in post #5.
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