c - how to print a matrix with a negated row more efficiently

Question

I wrote the following function:

void negate_row(const int n, const int r, int *a)
{
    if (r == 0)
    {
        printf("Matrix with negated row: ");
        printf("
");
        for (int y = 0; y < 3; y++)
        {
            *(a + 3 * r + y) = *(a + 3 * r + y) *(-1);
            printf("%d ", *(a + 3 * r + y));
        }
        printf("
");

        for (int y = 0; y < 3; y++)
        {
            *(a + 3 * r + y) = *(a + 3 * r + y) *(-1);
            printf("%d ", *(a + 3 * (r + 1) + y));
        }
        printf("
");

        for (int y = 0; y < 3; y++)
        {
            *(a + 3 * r + y) = *(a + 3 * r + y) *(-1);
            printf("%d ", *(a + 3 * (r + 2) + y));
        }
        printf("
");
    }

So basically what happens here is my function takes in a n X 3 matrix and negates a specific row using pointer arithmetic. I have been able to achieve that, I've also been able to figure out how to print that same matrix with the negated row. It's just the way I'm doing it is not efficient at all. I'd have to write an if statement for each row, ex if r == 0,1,2,3,4 etc... is there any way I can do this more efficiently?

Some clarifications: const int n decides the size of the matrix (n x 3), const int r decides what row is negated (0 <= r < n).

question from:https://stackoverflow.com/questions/66067240/how-to-print-a-matrix-with-a-negated-row-more-efficiently

Answer 1

A second loop will help. I generally find pointer code a bit harder to read. Particularly with Matrix manipulation you might be better off using array syntax instead of pointer syntax.

for (int y = 0; y < 3; y++)
{
    for (int x = 0; x < 3; x++)
    {
        printf("%d ", *(a + 3 * (r + x) + y));
    }

    printf("
");
}