Question

Out of curiosity, I change the line to set_array(&array1[0]) in the following from set_array[array1], the parameter is not the same type, but it works, any idea?

#include <stdio.h>
void set_array(int array[][9]);
int main(void) {
    int array1[4][9];
    for(int i = 0; i < 4; i++) {
        for(int j = 0; j < 9; j++) {
            array1[i][j] = j + 1;
        }
    }
    set_array(&array1[0]);
    for(int i = 0; i < 4; i++) {
        for(int j = 0; j < 9; j++) {
            printf("%d ", *(*(array1 + i) + j));
            //printf("%d ", array1[i][j]);
        }
        puts("\n");
    }
    return 0;

}
void set_array(int array[][9]) {
    for(int i = 0; i < 4; i++) {
        for(int j = 0; j < 9; j++) {
            array[i][j] = 1;
        }
    }   
};

Answer 1

here array1 is having the address of whole array of 1st row(or column)depending upon implementation. (&array1[0])is same as (array1) of type ()[n] sice array name itself contains the address of the ist element .But when you pass (&array) which of type ()[m][n] and compiler will throw an error because mismatch of argument type.

I think you haven’t read my comments in the last question you asked