Time Difference

[louvelle]

Hi guys.
I can't seem to find a proper code for getting the time difference.

This is the flow of the program:
1.) Input Time-in. (Should be in "0000" format. Example: "0800" [stands for 8:00 AM] )
2.) Input Time-out. (Example: "1700" [stands for 5:00 PM])
3.) Output the number of hours. (8 hours would be in this case)

The problem is with the minutes. In the flow of the code, i would normally go with "End - Start" and then divide with 100 to get the total number of hours. But if the first input is "759" (which stands for "7:59 AM") and the second input is "1702" (which stands for "5:02" PM), I always gets a result of "943" which is wrong and I'm kinda confused right now on how to get "902". Please help.

Thanks.

[akhileshbc]

I have made a pretty quick example in VB6:

Option Explicit
 
Private Sub Command1_Click()
    '~~~ Sample values
    Dim strStart As String
    Dim strEnd As String
       
    strStart = "759"
    strEnd = "1702"
    
    
    '~~~ Split the hours and minutes of both times
    Dim lngMin1 As Long
    Dim lngMin2 As Long
    
    Dim lngHr1 As Long
    Dim lngHr2 As Long
    
    '~~~ Minutes
    lngMin1 = CLng(Right$(strStart, 2))
    lngMin2 = CLng(Right$(strEnd, 2))
    
    '~~~ Hours
    lngHr1 = CLng(Mid$(strStart, 1, Len(strStart) - 2)) '~~~ will take both "759" as well as "0759"
    lngHr2 = CLng(Mid$(strEnd, 1, Len(strEnd) - 2))
    
    
    '~~~ Find the total minutes of each time
    Dim lngTime1 As Long
    Dim lngTime2 As Long
    
    lngTime1 = (lngHr1 * 60) + lngMin1
    lngTime2 = (lngHr2 * 60) + lngMin2
    
    '~~~ Subtract the total minutes
    Dim lngDiff As Long
    lngDiff = lngTime2 - lngTime1
    
    '~~~ Now we got the difference. So, divide the minutes by 60 (and take floor() of it). This will give the minutes.
    '~~~ ... and by taking MOD 60, you'll get the minutes
    Dim strTimeDiff As String
    strTimeDiff = CStr(Fix(lngDiff / 60)) & Format(CStr(lngDiff Mod 60), "0#") '~~~ "903"
    
    '~~~ Display the time difference
    MsgBox strTimeDiff
End Sub

I don't have C++ at the moment and it's a bit rusty. And, I hope you would be able to understand the VB6 code.

It's very simple code. And I hope you would be able to easily translate it to C++ code, once you get the idea.

[louvelle]

Thanks for the quick reply. I'll see what I can do.
I too am not familiar with C++.
Thanks again.

[anuvab1911]

#include<iostream>
using namespace std;
main()
{
struct tm   //defines a structure for holding the hour and minute of he input and output
{
int hh;
int mm;
};
int inp,outp,res,a,b,c,d;
tm tm1,tm2,tm3;
cout<<"Enter the start time in the HHMM format"<<endl;
cin>>inp;   //say 759
cout<<"Enter the end time in the HHMM format"<<endl;
cin>>outp;   //say 1702
tm1.mm=inp%100;   //59
tm1.hh=inp/100;    //7
tm2.mm=outp%100;   //2
tm2.hh=outp/100;   //17
tm3.hh=tm2.hh-tm1.hh;  //10
if((tm2.mm-tm1.mm)<0)  //As in your case 2-59=-57
{
tm3.mm=60+(tm2.mm-tm1.mm);  //so this becomes 3
tm3.hh--;    //decrease hour by one
}
else
tm3.mm=tm2.mm-tm1.mm;
cout<<tm3.mm+tm3.hh*100<<endl;    //outputs 903 to screen
}