C Help Magic Square

[winnerpl]

Hey guys I'm trying to get a magic square but I'm stuck with it printing
out only 1's and 0's in random places. Hope you experts can can provide
some input on what I'm doing wrong.

#include <stdio.h>
#define MAX 20

int x[MAX][MAX];

int size_request(void); //asks the user to enter the size x size square.
void fill_square(int[][], int); //fills up the magic square with the
appropriate numbers.
void print_square(int[], int); //prints out the magic square.

int main(void)
{

int roww;
int size = size_request();
x[size][size];

fill_square(x, size);

for(roww = 0; roww < size; roww++)
print_square(x[roww], size);

return 0;
}

int size_request(void)
{
int size;
do
{
printf("Enter a positive, odd integer in the range from 1 to 15:
");
scanf("%d", &size);
} while((size % 2 != 1) ||(size > 15) || (size < 1));

printf("\n");

return size;
}

void fill_square(int x[][MAX], int size)
{
int count;
int row = size - 1;
int col = row / 2;
int maxnum = size * size;

for(count = 1; count < maxnum; count++)
{
x [row][col] = 1;

if(x [row-2][col+1] == 0)
{
row = row - 2;
col = col + 1;
if (col == size)
col = col - size;
else if (row < 0)
row = row + size;
}
else
{
row --;
if (row < 0)
row = row + size;
}
}
return;
}

void print_square(int x[], int size)
{
int colw;

for(colw = 0; colw < size; colw++)
printf("%4d", x[colw] * x[colw]);

printf("\n");
return;
}



Thanks in advance guys.

 

[Jack Klein]

Hey guys I'm trying to get a magic square but I'm stuck with it printing
out only 1's and 0's in random places. Hope you experts can can provide
some input on what I'm doing wrong.

OK, first of all you are writing a program in some language that sort
of resembles C, but is not actually C.

#include <stdio.h>
#define MAX 20

int x[MAX][MAX];

int size_request(void); //asks the user to enter the size x size square.
void fill_square(int[][], int); //fills up the magic square with the
appropriate numbers.

Note, don't ever use // comments in C or C++ code posted to usenet.
Line wrapping by your news software has moved "appropriate numbers."
to the next line, where it is nothing but a syntax error when somebody
tries to compiler your code. Use /* */ comments, and line breaking
can't mess them up.

In any case, the prototype above is not valid C, it is a syntax error.
There is no such thing as a function array parameter with more than
one empty '[ ]' pair. If your compiler accepts this, it is either not
a C compiler or you are not invoking it to operate in standard C
conforming mode.

void print_square(int[], int); //prints out the magic square.

int main(void)
{

int roww;
int size = size_request();
x[size][size];

What do you think that this line does? You have already defined the
array at file scope. It doesn't change the size of the array. All it
does is evaluate the value of the int in the array at x[size][size],
which has the value 0, and then throws away that 0 value.

fill_square(x, size);

for(roww = 0; roww < size; roww++)
print_square(x[roww], size);

return 0;
}

int size_request(void)
{
int size;
do
{
printf("Enter a positive, odd integer in the range from 1 to 15:
");
scanf("%d", &size);
} while((size % 2 != 1) ||(size > 15) || (size < 1));

printf("\n");

return size;
}

void fill_square(int x[][MAX], int size)
{
int count;
int row = size - 1;
int col = row / 2;
int maxnum = size * size;

Let's see, if 'size' is 1, a value accepted by your size_request()
function, then 'row', 'col' are both 0 and 'maxnum' is 1.

for(count = 1; count < maxnum; count++)

In that case, this loop has a real problem, it will never execute at
all because 'count' starts at 1 and is never less than 'maxnum' if
'size' is 0.

{
x [row][col] = 1;

The line above is the only line that ever writes a value into the
array. The only value it ever writes is 1, so it is hard to see how
the array can have any values other than 0 (what it was initialized
with) or 1 in it. Perhaps you meant to write:

x [row][col] = count;

if(x [row-2][col+1] == 0)

Using 'row-2' as a subscript will produce undefined behavior if 'size'
is 1, although since this code won't execute when 'size' is 1, I don't
suppose that matters much.

{
row = row - 2;

Why not 'row -= 2'?

col = col + 1;

Why not 'col += 1' or just '++col'?

if (col == size)
col = col - size;

The line above could be replaced with 'col = 0'.

else if (row < 0)
row = row + size;
}
else
{
row --;
if (row < 0)
row = row + size;
}
}
return;
}

void print_square(int x[], int size)
{
int colw;

for(colw = 0; colw < size; colw++)
printf("%4d", x[colw] * x[colw]);

printf("\n");
return;
}



Thanks in advance guys.

Don't they teach people how to debug anymore? Have you single stepped
through your 'fill_square()' function to see what it does?

I made the following changes to your code:

1. Change prototype to 'void fill_square(int[MAX][MAX], int);' and
changed the function definition to match.

2. Eliminated the do-nothing 'x[size][size];' statement from main().

3. Changed 'x [row][col] = 1;' in fill_square() to 'x [row][col] =
count;'.

4. Added line right after that:

printf("x [%d][%d] = %d\n", row, col, count);

Now when I run your program and enter '3' for size, I get this output:

Enter a positive, odd integer in the range from 1 to 15: 3

x [2][1] = 1
x [0][2] = 2
x [-2][0] = 3
x [-1][1] = 4
x [0][2] = 5
x [-2][0] = 6
x [-1][1] = 7
x [0][2] = 8
0 0 64
0 0 0
0 1 0

....so it appears that you are using the expression [row-2] when row is
0, and you are accessing memory outside the array. You need to work
on your subscript expressions.

 

[Peter Nilsson]

void fill_square(int[][], int);

... the prototype above is not valid C,
Agreed.

it is a syntax error.

I guess you could argue your case here, given that the (outer) array
element's type can be deduced as incomplete from the syntax alone. [But a
part of me thinks you're stretching it with this statement. :-]

There is no such thing as a function array parameter with more than
one empty '[ ]' pair. If your compiler accepts this, it is either not
a C compiler or you are not invoking it to operate in standard C
conforming mode.

% dir /B decl.*
decl.c

% type decl.c
void blah(int [][]);

% gcc -c -ansi -pedantic decl.c
decl.c:1: warning: array type has incomplete element type

% dir /B decl.*
decl.c
decl.o

%

A conforming implementation _is_ allowed to accept the code, so long as it
emits the required diagnostic.

 

[Tim Rentsch]

void fill_square(int[][], int);

[some snippage]

In any case, the prototype above is not valid C, it is a syntax error.
There is no such thing as a function array parameter with more than
one empty '[ ]' pair. If your compiler accepts this, it is either not
a C compiler or you are not invoking it to operate in standard C
conforming mode.

Both a prototype and a function definition with an 'int[][]' type were
accepted by 'gcc -(ansi|std=c99) -pedantic' -- some warnings, but no
errors, and certainly no syntax errors. Is gcc in error here?
Or does the standard actually allow this?

Incidentally, the meaning assigned seems to be the same as though
it were like this:

int f( int (*)[] ); /* or int f( int [][] ); */

int f( int (*x)[] ){ /* or int f( int x[][] ){ */
return (*x)[0];
}

extern int a[];

int main(){
return f( &a ) == 0;
}

Of course, I'm making no claim that code like this is sensible.
Only asking if it's legal.

 

[Flash Gordon]

void fill_square(int[][], int);

[some snippage]

In any case, the prototype above is not valid C, it is a syntax
error. There is no such thing as a function array parameter with
more than one empty '[ ]' pair. If your compiler accepts this, it
is either not a C compiler or you are not invoking it to operate in
standard C conforming mode.

Both a prototype and a function definition with an 'int[][]' type were
accepted by 'gcc -(ansi|std=c99) -pedantic' -- some warnings, but no
errors, and certainly no syntax errors. Is gcc in error here?
Or does the standard actually allow this?

<snip>

The C standard does not discriminate between warnings and diagnostics,
and once a compiler has produced the required diagnostic it is allowed
to produce a program. So yes, gcc is allowed to do this since it has
produced a diagnostic.