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

This is a discussion on Help with writing a program that creates an odd n*n magic square where n is... within the C Programming forums, part of the General Programming Boards category; I am in a beginning C programming class and I have this assignment: Write a program that creates an N*N ...

  1. #1
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    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 few left that I cannot figure out. I understand that my main problem is visual studio doesn't support variable array size so I tried to compensate for this, but clearly I failed. I would really appreciate any help because this program is really frustrating me!

    Thank you!

    PHP Code:
     error C2057expected constant expression1 error C2466cannot allocate an array of constant size 0error C2057expected constant expression error C2466cannot allocate an array of constant size 0 error C2087'magic_square' missing subscripterror C2133'magic_square' unknown sizeerror C2143syntax error missing ';' before ')'error C2143syntax error missing ';' before ')' 

    Code:
    #include <stdio.h>#define array_size 15
    
    
    int main (void) 
    {
    	int row, column, row1, column1, 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 {
    
    
    		int magic_square [size][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)
    				col = 0;
    			if (magic_square [row][column] != 0) {
    				if (++row > (size -1))
    					row = 0;
    				if (--column < 0)
    					col = 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]);
    	}
    	}
    
    
    	return 0;
    }

  2. #2
    qny
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    Easy solution: just define the array with a big enough size (15x15) and ignore the unused elements if the user asks for a smaller one.

  3. #3
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    for (n, n<= size*size, n++) You sure you want to do this? Just a suggestion, but start with a 3X3 array to see if you have the code working properly

  4. #4
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    I thought I did define the array with my header: #define array_size 15

  5. #5
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    Okay actually I think I understsand what you're saying. I changed my magic_square [size][size] to magic_square [array_size][array_size], but I still have this error twice: error C2143: syntax error : missing ';' before ')'

  6. #6
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    Hint: Refer to post #3 above.

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