Output of a 2-d array needs to be understood

Question

This is the code https://ide.geeksforgeeks.org/8bYOzDDC9U to run this

#include <stdio.h>

char *c[] = {"GeksQuiz", "MCQ", "TEST", "QUIZ"};
char **cp[] = {c+3, c+2, c+1, c};
char ***cpp = cp;

int main()
{
    cpp++;
    printf("%s", cpp[0][0]);  // TEST
    printf("%s ", cpp[-1][-1]); // TEST
    return 0;
}

The output of both the printf() is TEST. Why is it so? When I do cpp++, the pointer moves to c+2. SO I understand that cpp[0][0] will be at the starting of "TEST" and that's why it is printing TEST. Can someone explain me for cpp[-1[-1]?

In C++ you should not assign a string literal to `char*` https://stackoverflow.com/questions/59670/how-to-get-rid-of-deprecated-conversion-from-string-constant-to-char-warnin — Cory Kramer, Apr 16 '20 at 14:57
`cpp[-1][-1]` is Undefined Behavior because `-1` is outside of the bounds of the array. — 0x5453, Apr 16 '20 at 14:57
Also I don't know if you are used to Python, but in C indexing `[-1]` is not the back of the array, if that's what you are trying to do — Cory Kramer, Apr 16 '20 at 14:57
btw this is something you need to understand if you want to become a [three-star-programmer](https://wiki.c2.com/?ThreeStarProgrammer), for everything else you should rather write code that is easier to write, read and understand — 463035818_is_not_an_ai, Apr 16 '20 at 14:58
@AkhileshGangwar Being a three-star programmer is generally a bad thing. — Thomas Jager, Apr 16 '20 at 15:03
don't try to learn or get inspiration from that that geeky site. I don't know how it is for C, but whenever I read there something on C++ i am shocked. — 463035818_is_not_an_ai, Apr 16 '20 at 15:03
@0x5453: `cpp[-1][-1]` has no undefined behavior. Pointer arithmetic with negative integers is defined by the C standard as long as the results point to elements of arrays or one beyond the end of an array, and these operations do. — Eric Postpischil, Apr 16 '20 at 15:04
@idclev463035818 I am trying to get my hands on pointers. SO I tried to solve some questions. And I stuck there. — ABD, Apr 16 '20 at 15:04
@EricPostpischil Good catch, I missed the `cpp++` above that. — 0x5453, Apr 16 '20 at 15:08

score 2 · Accepted Answer · answered Apr 16 '20 at 15:03

char ***cpp = cp; initializes cpp to point to the first element of cp, that is, cp[0].

After cpp++, cpp points to cp[1]. Then cpp[-1] refers to cp[0].

cp[0] contains c+3, which is a pointer to c[3]. So cp[0][-1] refers to c[2].

c[2] contains "Test", in the sense that it is a pointer to the first element of "Test", so printing it prints "Test".

Vlad from Moscow · Answer 2 · 2020-04-16T15:11:08.853

Initially the pointer cpp points to the first element of the array cp.

char ***cpp = cp;

After this statement

cpp++;

the pointer cpp points to the second element of the array cp. So this expression

cpp[-1]

gives the first element of the array cp

Pay attention to that the expression cpp[-1] is calculated like

*( cpp - 1 )

Thus in fact these two statements

cpp++;

and

cpp[-1]

may be considered like *( ( cpp += 1 ) - 1 ) that has the same effect if to write cpp[0] without incrementing the pointer cpp.

This (first ) element of the array cp. contains the value c+3 that is the pointer to the last element of the array c.

So cpp[-1][-1]give the element before the last element of the array c that is the pointer to the first character of the string literal "TEST".

Thus the string literal is outputted by this call of printf

printf("%s ", cpp[-1][-1])

Also bear in mind that in the presented code there is neither 2-d array. All arrays, c and cp, are one-dimensional arrays of pointers.

How can we consider cpp++ and cpp[-1] as cpp[0] ? This is something I don't understand in terms of pointers. — ABD, Apr 17 '20 at 05:05
@AkhileshGangwar Consider x + 1 - 1. What is the result? It is in fact the same as cpp++ - 1 because the result of cpp++ is cpp + 1. — Vlad from Moscow, Apr 17 '20 at 07:49

Output of a 2-d array needs to be understood

2 Answers2