2d array question

[ashlee]

main(void)
{
int i, j, k, m, n, p[2][3], q[2][3], r[2][3];
for(i = 0; i < 2; ++i)
for(j = 0; j < 3; ++j)
{
m = i + 3;
n = j + 2;
k = m%n;

p[i][j] = i + j;
q[i][j] = k;
}
for(i = 0; i < 2; ++i)
for(j = 0; j < 3; ++j)
printf("%i, %i\n", p[i][j], q[i][j]);}

so i see when the loop starts i = 0 j = 0 making the output p=0 q = 1
then when the loop is i=1 j = 0 i see the output being 1 and 0
but the next line of output is p=2 q=3 when i = 1 and j = 1. p = i+j which is 2... okay. k is 4%3 which is 1? not 3?

output:
0,1
1,0
2,3
1,0
2,1
3,0
maybe im not getting this right..

[new_ink2001]

Ashlee,

The result of the statement: a % b is the 'remainder' term in the expression:

a mod b = r if and only if a = n*b + r for some integer numbers n and 0 <= r < b

You may verify that 4 % 3 <==> 4 = 1*3 + 1, so you got that part right. According to the Wikipedia article, Modulo operation - Wikipedia, the free encyclopedia, "...some programming languages, such as C89, don't even define a result if either of n or a is negative." So be weary of negative operands....

What do those array's look like after the first set of loops, Ashlee? I got:

p = { {0, 1, 2} {1, 2, 3} }

and then for q, I did some substitution:

(i + 3) % (j + 2) is on the interval...
<==> [ 3 % (0 + 2), 3 % (2 + 2) ] when i is zero, and
<==> [ 4 % (0 + 2), 4 % (2 + 2) ] when i is one,
thus:

q = { {1, 0, 3}, {0, 1, 0} }

So the second set of loops produce:

0,1 -
1,0 | Section where 0 == i, and j is on [0,2]
2,3 -
1,0 -
2,1 | Section where 1 == i, and j is on [0, 2]
3,0 -

Did I loose you? Using interval notation for correctness proofs can be tedious, but it can clarify a suspected bug rather quickly. I hope this helps.

Best Regards,

New Ink -- Henry