指向2D数组的指针作为c中函数的参数

Question

I'm writing a Sudoku solving program and I'm a bit confused on using pointers with 2D arrays. Currently I am defining a puzzle like so:

int puzzle[9][9] = {
    {0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0}
};

and then I have a function called backtrack which will solve this recursively using backtracking. I need to pass puzzle to backtrack so I can modify it inside the function. Currently backtrack looks like

int backtrack(int (*p)[9][9]){
    int puzzle[9][9] = *p;

    //actual logic is here

    return 0;
}

but gcc gives an error saying that int puzzle[9][9] = *p; is an invalid initializer

I am under the impression that int (*p)[9][9] is a pointer to a 9 x 9 int array, so I should be able to turn this into a int puzzle[9][9] by dereferencing p , but this doesn't seem to work.

Answer 1

It is not possible to assign to all elements of an array at once using an assignment expression, it is possible to initialize some or all elements of an array when the array is defined.

A 1D array decays to a pointer. However, a 2D array does not decay to a pointer to a pointer.

If you declare backtrack like this:

int backtrack(int arr[][9]);

or even better

int backtrack(int r, int c, int arr[][c]);

and call like:

 int backtrack(puzzle);
 int backtrack(9,9, puzzle);

any modifications to the arr[x][y ] element modify the original array puzzle .

arr , the argument in backtrack , is of type int (*)[c] .

Edit: The explicit use of pointers in the calling function is of course possible as shown:

#include <stdio.h>
#include <stdlib.h>

#define NR_OF_ROWS  9

void backtrack1(int nr_of_columns, int *array){
    // access to 
    // array[i][j] =
    // *(array + i*nr_of_columns + j)
}

void backtrack2(int nr_of_columns, int array[][nr_of_columns]){
    //...
}

int main(void)
{
    int nr_of_columns = 9;       // number of columns      
    int *ptr1;                   // (to show how to init a pointer to puzzle1)    
    int (*ptr2)[nr_of_columns];  // (to show how to init a pointer to puzzle2)   

    int puzzle1[NR_OF_ROWS][nr_of_columns]; // declare puzzle1 
    int puzzle2[NR_OF_ROWS][nr_of_columns]; // declare puzzle2 

    ptr1 = &puzzle1[0][0];       // pointer `ptr1` points to first element in the puzzle1
    ptr2 = puzzle2;              // pointer `ptr2` points to first element in the puzzle2

    // 1a.  call `backtrack1` function
    backtrack1(nr_of_columns, ptr1); // or  `backtrack1(nr_of_columns,  &table1[0][0]);`

    // 2a. call `backtrack2` function  
    backtrack2(nr_of_columns, ptr2); // or simply `backtrack2(nr_of_columns, table2);

    return 0;     
}

Answer 2

Though the parameter declaration is correct but it seems you mean the following

int backtrack(int (*p)[9]){

and the function is called like

backtrack( puzzle );

When an array is used as an argument it is implicitly converted to pointer to its first element. In case of the array puzzle the first element of the array is the first "row" that is a one-dimensional array of the type int[9] .

The syntax to access an element of the array is look like

p[i][j]

where i and j are some indices.

Take into account that when the parameter is declared like

int backtrack(int p[9][9]){

nevertheless the compiler adjusts it in any case like

int backtrack(int (*p)[9]){

You could declare the parameter like

int backtrack(int (*p)[9][9] ){

but in this case the function is called like

backtrack( &puzzle );

and the body of the function would be too complicated. For example the syntax to access an element of the original array will look like

( *p )[i][j]

Also this statement in the function

int puzzle[9][9] = *p;

does not make sense. You may not initialize an array such a way. I think this statement is redundant in implementation of the function. You already have the poinetr p that points to the first "row" of the array. And you can change the original array or traverse its elements using this pointer. In fact all elements of the array are passed by reference through the pointer p.