Pointers dereferencing differences between [] and *

Question

I have come across an opinion that in case we have a pointer to a pointer, we can dereference it via [][] in case the allocated memory is adjacent. In case its not, only dereferencing via pointers arithmetic can be used. But I cannot quite reproduce that situation, i.e. here is an example code:

#include <stdio.h> #include <stdlib.h> #define ITEMS_NO 3 #define LETTERS_NO 4 int main(void) { char* theArray[ITEMS_NO]; char* dummyArr[ITEMS_NO]; for (int i = 0; i < ITEMS_NO; ++i) { theArray[i] = malloc(LETTERS_NO); dummyArr[i] = malloc(LETTERS_NO); // make sure mem is not adjecent for (int k = 0; k < LETTERS_NO - 1; ++k) { theArray[i][k] = '0' + i; dummyArr[i][k] = 'z'; } theArray[i][LETTERS_NO - 1] = 0; dummyArr[i][LETTERS_NO - 1] = 0; } for (int i = 0; i < ITEMS_NO; ++i) { printf("item by square bracket %i = %c \n", i, theArray[i][1]); printf("item by ptr %i = %c \n", i, *((*(theArray + i)) + 1)); free(theArray[i]); free(dummyArr[i]); } }

In this example i allocate the dummyArr in order to make sure that the allocated memory for the theArray is not adjacentnext to each other. Both printf calls give the same results. Am I not getting something here or do both dereferencing methods work exactly the same way?

The C language itself defines that p[i] and *(p + i) are exactly identical and interchangeable with each other; whatever alignment concerns there are work the same way either way. — Steve Friedl
– Steve Friedl, Commented Dec 11, 2019 at 14:53
* and [] is the same depending on the type of pointer it is. For instance if the pointer is of type int then pointer[1] is sizeof(int) passed the first index. Likewise *(pointer+1) is also sizeof(int) passed the first index. And I'm not exactly sure how you're checking the memory isn't alligned. If you want to target a specific byte after index 0 you can cast it to char then add your offset. — Tofu
– Tofu, Commented Dec 11, 2019 at 14:54

dbush · Accepted Answer · 2019-12-11 14:55:02Z

The two methods are exactly equivalent. Section 6.5.2.1p2 of the C standard regarding array subscripting states:

A postfix expression followed by an expression in square brackets [] is a subscripted designation of an element of an array object. The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2))). Because of the conversion rules that apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the initial element of an array object) and E2 is an integer, E1[E2] designates the E2-th element of E1 (counting from zero).

The fact that you have a pointer-to-pointer, where each pointer points to a distinct array, as opposed to a true 2D array doesn't change this.

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