Sum of Range

[ccorkran]

Hello everyone,
I am writing a function to find a sum of a range. I have a function that works in the main function, but I cannot figure how to call the function in main.

this is what I have.

  #include <iostream>

using namespace std;

int sumOfRange(int, int);

int main() {
	int sumOfRange(int firstName, int lastName);
	return ;
} // End of main

int sumOfRange(int firstNum, int lastNum) {
	int Sum = 0;
	int First, Last;

	cout << "Enter the first: ";
	cin >> First;
	cout << "Enter the last:  ";
	cin >> Last;

	if (First > Last)
	{
		int Temp;

		Temp = First;
		First = Last;
		Last = Temp;
	}

	for (int Counter = First; Counter <= Last; Counter++)
		Sum += Counter;

	cout << Sum << endl;

	return 0;
}

[DonnaPin]

Hello, I am just a beginner too, so what I say might very well be wrong, but I think your problem might be that you are not declaring your paramters ofsumOfRrange before you pass them. I have your program working as this

  #include <iostream>

using namespace std;

int sumOfRange(int, int);

int main() {
    int firstName;
    int lastName;
	sumOfRange(firstName, lastName);
	return 0;
} // End of main

int sumOfRange(int firstNum, int lastNum) {
	int Sum = 0;
	int First, Last;

	cout << "Enter the first: ";
	cin >> First;
	cout << "Enter the last:  ";
	cin >> Last;

	if (First > Last)
	{
		int Temp;

		Temp = First;
		First = Last;
		Last = Temp;
	}

	for (int Counter = First; Counter <= Last; Counter++)
		Sum += Counter;

	cout << Sum << endl;

	return 0;
}

[Ganado]

Yep you're right. By declaring the variables inside the [what would be a] function call, it actually makes it a function declaration: int sumOfRange(int firstName, int lastName); declares a function (redundant because it's already declared on line 5) but does not call it.

[Handy Andy]

Hello ccorkran,

Even with DonnaPin's suggestions this is what I see:

#include <iostream>

using namespace std;

int sumOfRange(int, int);  // <--- The variable names may not be required, but they do help.

int main()
{
    int firstName;  // <--- What does Name have to d with a numeric variable?
    int lastName;  // <--- Uninitialized and contain garbage values.

    sumOfRange(firstName, lastName);  // <--- Uninitialized variables containing garbage values. Also an error with
                                      // my compiler. Gives a warning in the shell program.

    return 0;  // <--- Not required, but makes a good break point.
} // End of main

// <--- This should be a void vunction.
int sumOfRange(int firstNum, int lastNum)  // <--- Variables received and with a better name.
                                           // Still contain garbage values and never used int the function.
{
    int Sum = 0;
    int First, Last;

    cout << "Enter the first: ";
    cin >> First;

    cout << "Enter the last:  ";
    cin >> Last;

    if (First > Last)
    {
        int Temp;

        Temp = First;
        First = Last;
        Last = Temp;
    }

    for (int Counter = First; Counter <= Last; Counter++)  // <--- Loops 1 more time with the "<=". Is that what you want?
        Sum += Counter;  // <--- Sum may be greater than what you expect.

    cout << Sum << endl;

    return 0;  // <--- This is not needed as there is nothing to return. And if
               // you make it a void function you can not return a value.
}

Andy

[ccorkran]

Thank you both, it will run in the stacks on this website, but will not run in visual studio because it says there are unused variables on line 8 and i have no clue what is going on.

#include <iostream>

using namespace std;

int sumOfRange(int firstNum, int lastNum);

int main() {
	int firstNum, lastNum;
	cout << sumOfRange(firstNum, lastNum);
	/*int sumOfRange(int firstNum, int lastNum);*/
	
} // End of main

int sumOfRange(int firstNum, int lastNum) {
	int Sum = 0;
	int First, Last;

	cout << "Enter the first: ";
	cin >> First;
	cout << "Enter the last:  ";
	cin >> Last;

	if (First > Last) {
		int Temp;

		Temp = First;
		First = Last;
		Last = Temp;
	}

	for (int Counter = First; Counter < Last; Counter++)
		Sum += Counter;

	cout << Sum << endl;

	return 0;
}

[deleted account xyzzy]

#include <iostream>

void sumOfRange();

int main()
{
   sumOfRange();
}

void sumOfRange()
{
   std::cout << "Enter the first: ";
   int first;
   std::cin >> first;

   std::cout << "Enter the last:  ";
   int last;
   std::cin >> last;

   if (first > last)
   {
      int temp = first;
      first    = last;
      last     = temp;
   }

   int sum { };

   for (int counter = first; counter <= last; counter++) sum += counter;

   std::cout << "\nThe sum of " << first << " to " << last << " is: " << sum << '\n';
}

Enter the first: 1
Enter the last:  100

The sum of 1 to 100 is: 5050

[lastchance]

#include <iostream>
#include <cstdlib>
using namespace std;

int sumOfRange( int firstNum, int lastNum )
{
   return ( ( abs( lastNum - firstNum ) + 1 ) * ( lastNum + firstNum ) ) / 2;
}

int main()
{
   int First, Last;
   cout << "Enter the two numbers: ";   cin >> First >> Last;
   cout << sumOfRange( First, Last ) << '\n';
}

Enter the two numbers: 100 1
5050

[Handy Andy]

Hello ccorkran,

you wrote:

but will not run in visual studio because it says there are unused variables on line 8 and i have no clue what is going on.

Strange my VS 2017 says:

Severity  Code	Description	Project	                        File           Line	
Error	C4700	uninitialized local variable 'lastNum' used	main v2.cpp	11	

If you did not understand the comments I put in the code that I posted just ask.

In "main" you are defining 2 variables "firstNum" and "lastNum", but they have no value just what happens to be in the memory they use for storage. Fo me this usually turns into an "int" of "-858993460" or known as garbage. Then when you try to call the function with out giving "firstNum" and "lastNum" a proper value the compiler complains.

You define your function as:

int sumOfRange(int firstNum, int lastNum) {
	int Sum = 0;
	int First, Last;

	cout << "Enter the first: ";
	cin >> First;
	cout << "Enter the last:  ";
	cin >> Last;

	if (First > Last) {
		int Temp;

		Temp = First;
		First = Last;
		Last = Temp;
	}

	for (int Counter = First; Counter < Last; Counter++)
		Sum += Counter;

	cout << Sum << endl;

	return 0;
}

You have sent "firstNum" and "lastNum" to your function, but where in this code do you use them.

If you do not understand just ask.

Andy

[ccorkran]

thank you Andy,
That helped a lot and is to clear to me now.

[Manga]

#include <iostream>

using namespace std;

int sumOfRange(int, int);

int main() {
    
    int First, Last;

	cout << "Enter the first: ";
	cin >> First;
	cout << "Enter the last:  ";
	cin >> Last;
	
	cout << sumOfRange(First, Last);
	
	return 0;
} // End of main

int sumOfRange(int First, int Last) {
    
     if (First > Last)
     {
     return ((First-Last+1)*(First+Last))/2;
    }
    else
    {
      return ((Last-First+1)*(First+Last))/2;  
    }

}

[seeplus]

@manga. Just use abs() like lastchance did in his code above.

Also, there's no need for the else as the last statement in the true condition is a return - and the tertiary operator can be used.

Without using abs():

#include <iostream>
#include <string>
#include <limits>

using namespace std;

int sumOfRange(int, int);
int getInt(const std::string& prm);

int main()
{
	const auto first {getInt("Enter the first: ")};
	const auto last {getInt("Enter the last: ")};

	cout << sumOfRange(first, last) << '\n';
}

int sumOfRange(int first, int last)
{
	return (((first > last ? first - last : last - first) + 1) * (first + last)) / 2;
}

int getInt(const std::string& prm)
{
	int i {};

	while ((std::cout << prm) && (!(std::cin >> i) || std::cin.peek() != '\n')) {
		std::cout << "Not an integer\n";
		std::cin.clear();
		std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
	}

	std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
	return i;
}

[seeplus]

The formula foe the sum of the range of consecutive integer numbers is given as

(n - m + 1) * (m + n) / 2

where m is the first number and n the last. This is derived from the standard formula for the sum of x consecutive numbers starting from 1 which is x * (x + 1) / 2

The required sum is the sum of the numbers from 1 to (m - 1) subtracted from the sum of numbers from 1 to n. Thus:

n * (n + 1) / 2 - (m - 1) * (m - 1 + 1) / 2
[n * (n + 1) - (m - 1) * m] / 2
[n* n + n - m * m + m] / 2
[(n + m) + n * n- m * m]  / 2
[(n + m) + (n - m) * (n + m)] / 2
(n + m)( n - m +1) / 2

[lastchance]

It's quicker just to do (as for any arithmetic progression):

sum = (number of elements) * (average element)
    = ( n - m + 1 ) * ( m + n ) / 2

(in the case where, WMLOG, n is the larger number )

[seeplus]

Yep. As from first principles:

S = m + (m + 1) + (m + 2) + ... + (n - 2) + (n - 1) + n
S = n + (n - 1) + (n - 2) + ... + (m + 2) + (m + 1) + m

S + S = (n + m) + (m + 1 + n - 1) + (m + 2 + n - 2) + ... + (n - 2 + m + 2) + (n - 1 + m + 1) + (n + m)

2 * S = (n + m) + (n + m) + (n + m) + ... for (n - m + 1) terms
2 * S = (n + m) * (n - m + 1)
S = (n + m) * (n - m + 1) / 2
S = (n - m + 1) * (n + m) / 2