如何使用指针符号访问2d数组中的结构元素？

Question

English is not my first language so I'm not sure if I formulated the question the right way. If anyone would like to correct me feel free to do so. Now,

I understand that arr[i][j] is the same as *(arr+i+j) but, now I have this

struct whatever{
int n;
int k;};
int main()
{
    struct whatever arr[4][4];
}

and I need to know how to access 'n' inside the structure via the pointer notation . I tried (*(arr+i+j).n)

(it just gives an error ': request for member 'n' in something not a structure or union)

and a whole lot of other variations and I can't find anything related on the forums for the past 2 hours.

Answer 1

*(arr + i + j) is the same as arr[i + j] .

( *(a + b) is a[b] – the above are also equivalent to (arr + j)[i] , i[j + arr] , and (i + j)[arr] . But don't go there.)

arr[i][j] is *(*(arr + i) + j) .
(Or *(j + *(i + arr)) . Don't do it.)

The sensible thing to do, if you're not taking part in an obfuscation contest, is to not use pointer arithmetic when you can avoid it.
Pointer arithmetic makes it very easy to write code that seems right, is completely wrong, and still compiles and then has to be debugged when you could have done something useful instead.

Answer 2

As I noted in a comment :

You'd need to write (*(*(arr + i) + j)).n — which is an excellent reason for using the arr[i][j].n notation. I had to build it up piecemeal from the inside out: *(arr + i) gives the result of arr[i] ; you then have to treat that as a pointer expression and give it the *(ptr + j) treatment; and then the precedence rules say . binds tighter than * , so you need the extra parentheses. Except as an academic exercise, don't write that using pointers and dereferences.

You can also omit the outer parentheses and * by using the arrow -> operator instead of the dot . operator: (*(arr + i) + j)->n .

Here's some code:

#include <stdio.h>

struct whatever{
    int n;
    int k;
};

int main(void)
{
    struct whatever arr[4][4] = { [2] = { [3] = { 371, 4 } } };
    printf("arr[2][3].n           = %d\n", arr[2][3].n);
    int i = 2;
    int j = 3;
    printf("(*(*(arr + %d) + %d)).n = %d\n", i, j, (*(*(arr + i) + j)).n);
    printf("(*(arr + %d) + %d)->n   = %d\n", i, j, (*(arr + i) + j)->n);
    return 0;
}

And the output is:

arr[2][3].n           = 371
(*(*(arr + 2) + 3)).n = 371
(*(arr + 2) + 3)->n   = 371

(The initialization uses C99 designated initializers to set element arr[2][3] to the value (struct whatever){ 371, 4 } — which is the compound literal notation for the element of type struct whatever ; compound literals are a C99 feature too.)

This code was compiled by GCC 7.2.0 on a Mac running macOS High Sierra 10.13.2.