指向多维数组的指针

Question

/* Demonstrates passing a pointer to a multidimensional */
/* array to a function. */

#include <stdio.h>

void printarray_1(int (*ptr)[4]);
void printarray_2(int (*ptr)[4], int n);

int main(void)
{
    int multi[3][4] = { {1, 2, 3, 4},
                        {5, 6, 7, 8},
                        {9, 10, 11, 12} };

    // ptr is a pointer to an array of 4 ints.
    int (*ptr)[4], count;

    // Set ptr to point to the first element of multi.
    ptr = multi;

    // With each loop, ptr is incremented tto point to the next
    // element (that is, the next 4-elements integer array) of multi.

    for (count = 0; count < 3; count++)
        printarray_1(ptr++);

    puts("\n\nPress Enter...");
    getchar();
    printarray_2(multi, 3);
    printf("\n");

    return 0;
}

void printarray_1(int (*ptr)[4])
{
    // Prints the elements of a single 4-element integer array.
    // p is a pointer to type int. You must use a typecast
    // to make p equal to the address in ptr.

    int *p, count;
    p = (int *)ptr;

    for (count = 0; count < 4; count++)
        printf("\n%d", *p++);
}

void printarray_2(int (*ptr)[4], int n)
{
    // Prints the elements of an n by 4-element integer arrray.

    int *p, count;
    p = (int *)ptr;

    for (count = 0; count < 4; count++)
        printf("\n%d", *p++);
}

In the definition of printarray_1 & 2 functions, the int pointer p is assigned to (int *)ptr. why?

In the pointer declaration of the main function, the parenthesis puts *ptr in a higher precedence than [4]. but (int *)ptr does not make sense to me. Would you please explain why?

Answer 1

The syntactically and semantically correct way to obtain p from ptr is:

    p = *ptr;

... as you are trying to extract an array int p[4] (declared as int *p ) from a pointer to an array int(*ptr)[4] . This eliminates the need for any casting.

Answer 2

(int *)p it the same like int *p
i advice you to make a simple program and try it