1 loop structure to find the sum of odd and even integers

saqib_ · #1 24-Oct-2009, 02:21

Hello, I am new to this forum and to programming. I was trying to make a small program that calculates the sum of odd and even integers up to the limit provided by the user. The code is given below. The problem is to find the even sum. I am stuck at this point. Please guide me.

CPP / C++ / C Code:

/* 
   This program calculates the sum of odd and even integers upto a given
   limit by the user using one loop structure only. 
*/

// ---> Start program

#include <iostream.h>
#include <conio.h>

int main()
{
    // ---> Declare variables and thier values
    
    int oddsum(0), num(1), evensum(0), userlim;
    
    // ---> Get user input
    
    cout << "Please enter the limit : ";
    cin  >> userlim;
    
    // ---> Calculate the sum of odd and even integers upto the limit
    
    // ---> Using while loop
    
    while(num<=userlim)
    {
                       oddsum+=num;
                       num+=2;
    }
    
    // ---> Show the result
    
    cout << "The sum of odd integers upto " << userlim << " is : " << oddsum << "\n";
    cout << "The sum of even integers upto " << userlim << " is : " << evensum;
    
    // End program
    
    getch();
    return 0;
    
}

Mexican Bob · #2 24-Oct-2009, 07:41

There is no standard header named conio.h...don't use it unless required by your course/instructor...and then question why you must use it.

I don't think that I'd use a while conditional; rather a for seems more suitable:

CPP / C++ / C Code:

#include <iostream>

using namespace std;

int main()
{
    int userlimit;
    cout << "Enter an integer limit: ";
    cin >> userlimit;

    if(userlimit < 2)
    {
	cout << "...not much to do!" << endl;
	exit(0);
    }

    int oddsum = 1;
    int evensum = 0;

    for(int i = 2; i <= userlimit; i++)
    {
	if(i % 2 == 0)
	{
	    evensum += i;
	}
	else
	{
	    oddsum += i;
	}
    }
    cout << "Sum of odd  integers between 1 and " << userlimit << " = " << oddsum << endl;	
    cout << "Sum of even integers between 1 and " << userlimit << " = " << evensum << endl;	

    return 0;
}

MxB

saqib_ · #3 24-Oct-2009, 08:00

Thank you for your guidance. You said that conion.h is not a standard header. I have used your code but the window disappears so fast that I cannot see the output. However if I use conio.h with getch(); in the end I can view my output of the program. Is it wrong to use it and write the code as you did in the example?

Mexican Bob · #4 24-Oct-2009, 09:13

Quote:

Originally Posted by saqib_

Thank you for your guidance. You said that conion.h is not a standard header. I have used your code but the window disappears so fast that I cannot see the output. However if I use conio.h with getch(); in the end I can view my output of the program. Is it wrong to use it and write the code as you did in the example?

"Wrong" is relative. There are alternatives to using non-standard libraries and headers:

CPP / C++ / C Code:

#include <iostream>

using namespace std;

int main()
{
    int userlimit;
    cout << "Enter an integer limit: ";
    cin >> userlimit;

    if(userlimit < 2)
    {
	cout << "...not much to do!" << endl;
	exit(0);
    }

    int oddsum = 1;
    int evensum = 0;

    for(int i = 2; i <= userlimit; i++)
    {
	if(i % 2 == 0)
	{
	    evensum += i;
	}
	else
	{
	    oddsum += i;
	}
    }
    cout << "Sum of odd  integers between 1 and " << userlimit << " = " << oddsum << endl;	
    cout << "Sum of even integers between 1 and " << userlimit << " = " << evensum << endl;	

    cout << "Type a key and then press Enter to continue...";
    char key;
    cin >> key;
    

    return 0;
}

MxB

Howard_L · #5 26-Oct-2009, 21:10

You can also pause with this without having to use a variable:

CPP / C++ / C Code:

  cout << "Press return to continue ";
  cin.get();

Mexican Bob · #6 27-Oct-2009, 04:11

Quote:

Originally Posted by Howard_L

You can also pause with this without having to use a variable:

CPP / C++ / C Code:

  cout << "Press return to continue ";
  cin.get();

I'm not sure that this is true in all cases on all platforms. For example, the following change:

CPP / C++ / C Code:

/*
    cout << "Type a key and then press Enter to continue...";
    char key;
    cin >> key;
    
*/
    cout << "Press return to continue...";
    cin.get();

...just ran without pausing, as shown below:

Code:

bash-3.2$ ./saqlib
Enter an integer limit: 200
Sum of odd  integers between 1 and 200 = 10000
Sum of even integers between 1 and 200 = 10100
Press return to continue...bash-3.2$

...on Mac OS X 10.5.8 using GNU GCC.

MxB

Kimmo · #7 27-Oct-2009, 08:12

Quote:

Originally Posted by Howard_L

You can also pause with this without having to use a variable:

CPP / C++ / C Code:

  cout << "Press return to continue ";
  cin.get();

The difference between this and what Bob showed is that this will always read the next character, whereas Bob's version discards whitespace. And when people do stuff like

CPP / C++ / C Code:

int anInt;
cout << "Enter an int: ";
cin >> anInt;

there will a newline waiting in the input stream ready to be read by cin.get(). I'm not saying Bob's solution is better in my opinion, just that

CPP / C++ / C Code:

cin.get();

is not equal to

CPP / C++ / C Code:

cin >> aChar;

Howard_L · #8 27-Oct-2009, 09:33

Ah yes, the leftover '\n' and extra characters typed. Good point M.
And nice explaination Kimmo.

Well I guess you could "flush" stdin after user input like you should do any time you want to get meaningful subsequent user input.

CPP / C++ / C Code:

    cin >> userlimit;
    while( (cin.get()) != '\n');    // empty remaining chars from cin (stdin)
    //blah blah
    cout << "Press return to continue ";
    cin.get();

That works for me , does that work on your platform too?
My platform is pine flooring on 2x4 joists about 12" above dirt... (gcc / fc6-linux)

Kimmo · #9 27-Oct-2009, 13:21

CPP / C++ / C Code:

    cin >> userlimit;
    while( (cin.get()) != '\n');    // empty remaining chars from cin (stdin)
    //blah blah
    cout << "Press return to continue ";
    cin.get();

Perhaps, but then you need to be sure every time there actually is a newline. An alternative is to use getline and stringstreams.

CPP / C++ / C Code:

#include <iostream>
#include <sstream>
#include <string>

int main()
{
    using std::cout;
    using std::endl;
    using std::cin;
    using std::istringstream;
    using std::string;
    
    int anInt(0);
    string input;
    
    cout << "Enter some input (a small integer could be a good idea): ";
    getline(cin, input);
    
    istringstream inputStream(input);
    inputStream >> anInt;
    
    cout << endl;
    
    if (!inputStream.fail() && inputStream.eof())
        cout << "Great." << endl;
    else
        cout << input << " is probably not an integer" << endl;
        
    cout << endl << "Press Enter to continue...";
        
    cin.get();
    
    return 0;
}

Meaningless output:

Code:

Enter some input (a small integer could be a good idea): Talo

Talo is probably not an integer

Press Enter to continue...

Howard_L · #10 27-Oct-2009, 14:27

Wow, that's an interesting way of handling user input.
You mean there is a way for user to enter data without pressing the Enter key?
Doesn't this always leave at least the '\n' like scanf()?

CPP / C++ / C Code:

cin << datatype;

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1 loop structure to find the sum of odd and even integers

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Re: 1 loop structure to find the sum of odd and even integers

Re: 1 loop structure to find the sum of odd and even integers

Re: 1 loop structure to find the sum of odd and even integers