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Magic Sqaure -- sum of diagonals help

Sunny101 (67)
Here's part of my code so far: 

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void main ()
{
	int magic [row] [col];

	//call functions
	getData( magic, row , col);
	double Determine = isMagic (magic, row, col);

}



void getData (int magic [] [col], const int row, const int col)
{
	cout << "Enter 4 numbers for each row. \n";
	for ( int row = 0; row < 4 ; row ++ )
	{
		cout << "\n";
		cout << "  Row " << row + 1 << endl;
		cout << "-----------\n";

		for ( int col = 0; col < 4 ; col ++)
		{
			cout << "Column " << col + 1 << ": ";
			cin >> magic [row] [col]; 
		}
	}
}


double isMagic (int magic [][col], const int row, const int col)
{
	int sum = 0;

	// Add rows
	for (int row = 0; row < 4 ; row ++)
	{
		sum = 0;

		for (int col = 0 ; col < 4 ; col ++)
		{
			sum += magic [row] [col];
		}
		
		cout << " The sum for row " << row + 1 << " is " << sum << endl;
	}

	//Add columns
	for (int col = 0; col < 4 ; col ++)
	{
		sum = 0 ;

		for (int row = 0 ; row < 4 ; row ++)
		{
			sum += magic [row] [col];
		}
		
		cout << " The sum for column " << col + 1 << " is " << sum << endl;
	}

	//Add Diagonal 1 
	
	

		for (int row = 0 ; row < 4 ; row ++)
		{
			sum += magic [row ] [];;
		}
		
		cout << " The sum for D1 is " << sum << endl;
	

    return sum;

}


I am not sure how to go about getting the sum of the diagonals. Any ideas?
Anthony Hernandez (53)
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for (int rc = 0; rc < 4; rc++)
{
    sum+= magic[rc][rc];
}


So it will add the diagonal in the middle:
0,0
1,1
2,2
3,3

* 0 0 0
0 * 0 0
0 0 * 0
0 0 0 *

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for (int rc =1; rc < 4; rc++)
{
    sum += magic[rc-1][rc];
}


0,1
1,2
2,3

0 * 0 0
0 0 * 0
0 0 0 *
0 0 0 0

I'm sure you can figure it out from here.
Sunny101 (67)
Wow! Thanks. So if I wanted to find the middle of the opposite side, it would look like this:

0,3
1, 2
2,1
3, 0

therefore making the code:
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for (int rc = 0; rc < 4; rc --)
{
    sum+= magic[rc][rc];
}


Sunny101 (67)
Actually nevermind. I'm guessing the change would have an addition interation and subtraction?

ps: Thanks Anthony
Anthony Hernandez (53)
I was thinking about it and I'm sure you do the same thing as

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for (int rc =1; rc < 4; rc++)
{
    sum += magic[rc-1][rc];
}


except you flip the sum part to this magic[rc][rc-1] for finding the diagonals below the middle one

And yes, three iterations in total since you need to solve above the middle, below the middle and the middle itself.
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