Thread: Help with writing a program that creates an odd n*n magic square where n is...

I am in a beginning C programming class and I have this assignment: Write a program that creates an N*N magic square, i.e., a square arrangement of the numbers 1,2,…,N^2 in which the sum of rows, columns, and diagonals are the same. The user will specify the size of the square matrix: N. The value N must be an odd number between 1 and 15.


I am using visual studio 2010. Also I am only allowed to use C89. My program will not compile and I have been trying to fix my problems, but I still have a problem with what the program prints out. My program prints out a vertical column of the magic square. For example if my magic square is size 3, it'll print:
8
1
6


3
5
7


4
9
2


Help please. Thank you!

Code:

#include <stdio.h>
#define array_size 15


int main (void) 
{
	int row, column, size, n = 2, magic_square [array_size][array_size];
	


	for ( ; ; ) {


	printf ("Enter size of magic square: ");


	scanf ("%d", &size);


	if (size % 2 == 0)
		printf ("Enter an odd number!\n");


	else if (size <= 0 || size > 15)
		printf ("Enter a number between 1 and 15!\n");


	else {


		magic_square [array_size][array_size];


		for (row = 0; row < size; row++) {
			for (column = 0; column < size; column++) {
				magic_square [row][column] = 0;
			}
		}


		row = 0;
		column = (size/2);
		magic_square [row][column] = 1;


		for (n; n<= size*size; n++) {
			if (--row < 0)
				row = (size -1);
			if (++column > size -1)
				column = 0;
			if (magic_square [row][column] != 0) {
				if (++row > (size -1))
					row = 0;
				if (--column < 0)
					column = size - 1;
				while (magic_square [row][column] != 0)
					if (++row > (size -1))
						row = 0;
			}


			magic_square [row][column] = n;
		}


				for (row = 0; row < size; row++) {
					for (column = 0; column < size; column++) {
						printf ("%4d\n", magic_square [row][column]);
					}
					printf ("\n");
				}


				break;
	}
	}
	return 0;
}

Sorry I figured it out!