How to modify this code to get the day number of a week and hours passed ?

Question

1.00/5 (1 vote)

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C++

#include <iostream>
    #include<cmath>
    #include<ctime>
    #include<string>
    #include <iomanip>
    #include <fstream>
  
    using namespace std;
    
    string today(int day, string todayName);
    int hourLeft(int hour);
    
    
    int main()
    {
    	int day,hour;
    	cin >> day>>hour;
    
    	string dayName;
    	cout << today(day, dayName) << endl;
    	cout << hourLeft(hour) << endl;
    }
    string today(int day, string todayName)
    {
    	switch (day)
    	{
    	case 1: return todayName = "sunday";
    		break;
    	case 2: return todayName = "monday";
    		break;
    	case 3: return todayName = "tuesday";
    		break;
    	case 4: return todayName = "wednesday";
    		break;
    	case 5:return  todayName = "thursday";
    		break;
    	case 6: return todayName = "friday";
    		break;
    	case 7:return  todayName = "saturday";
    		break;
    	}
    }
    int hourLeft(int hour)
    {
    	int theHourLeft = 24 - hour;
    	return theHourLeft;
    }

What I have tried:

Edit (CHill60): Blank line removed. See above for code

Posted 10-Nov-20 7:26am

mahmoud1998

Updated 11-Nov-20 13:37pm

v4

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Comments

Richard MacCutchan 10-Nov-20 15:05pm

What exactly is the problem?

KarstenK 11-Nov-20 12:50pm

The question is unclear.

jeron1 11-Nov-20 17:20pm

In addition to what Greg Utas posted, you should qualify your inputs. If the user types in '10' for the day variable or '19722' for your hour variable your code should handle it somewhere, otherwise you'll get operation you didn't expect or want. In many instances it is also good practice to have a default: statement in your switch statement to deal with data not handled by the case: statements.

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Greg Utas · Answer 1 · 2020-11-11T10:52:00

I can guess what's wrong. I'm not sure what

return todayName = "dayname";

is going to return, but my guess is that it won't be what you want. todayName doesn't need to be a parameter to this function. It already returns a string, so just write

return "dayname";

in all those places. You don't even need the break statements that follow, and your main function will just do

cout << today(day) << endl;

If you actually wanted to update the dayName local variable in main, you would have to declare your today function as

void today(int day, string& todayName)

Note the &, which tells the code to update the argument that is passed into the function instead of working on a copy which is lost when the function returns. With this approach, main would need to do

today(day, dayName);
cout << dayName << endl;