无法将数组传递给递归函数C ++

Question

When I run the function "check_row" within itself, there is a problem with the way I'm trying to pass in the array "sudoku_temp", but I'm not sure what I'm doing wrong. Am I missing something?

int check_row(int j_position, int generated_value, int sudoku_temp[][9]){
for (int i_position = 0; i_position < 9; i_position++)
{
    if (generated_value == sudoku_temp[i_position][j_position])
    {
        generated_value = generate_number();
        check_row(j_position, generated_value, sudoku_temp[][j_position]);
    }
    else
        return generated_value;

}

}

To clarify, the problem is when I try to call on the function within itself. Thanks.

Answer 1

use a variable instead of fixed length like 9. since in recursion the argument may be different.

int check_row(int j_position, int generated_value, int sudoku_temp[][n])
{
    for (int i_position = 0; i_position < n; i_position++)
    {
        if (generated_value == sudoku_temp[i_position][j_position])
        {
            generated_value = generate_number();
            check_row(j_position, generated_value, sudoku_temp[][j_position]);
        }
        else
            return generated_value;

    }
}

Answer 2

When you want to extract a value from an array, you cannot leave empty brackets like you did. On the other hand, the function prototype looks similar and it's perfectly fine. Why?

int sudoku_temp[][9] // A 2D array of integers second dimension of which is of size 9

Or in other words, an array (of unknown size) of arrays of size 9. We are not telling the compiler how big the array really is and we don't have to in this case as it is simply given to us as an argument.

When accessing elements on the other hand, we cannot leave empty brackets for a simple reason: we want to access an element and the compiler has to know which one. Cutting to the chase - removing the empty [] should solve your problem.