Allocating array in quadriple pointer

Question

Suppose I have int ****ptr;. I want to allocate a dynamic 1-dimensional array on the end of this ptr, so when I type ***ptr[4] I would yield an element. Please, help.

p.s. I am not using this code in my real applications, it's just an intellectual exercise to understand how pointers work. I cannot directly do ***ptr = malloc(sizeof(int)*size_of_arr)); right? BEcause this way i will not be able to yield any elements

Is this just an intellectual exercise? If you are actually *using* code like this, you are almost certainly going down the wrong path. — Beta, Sep 30 '15 at 22:50
Parenthesis to the rescue.. Have you tried to parenthesize your code? And perhaps break the process down into 2-3 lines of code, perhaps using some local helper variables? — Mike Dinescu, Sep 30 '15 at 22:50
No production level code would use such construct. Even asking such from an academical viewpoint, its pretty poorly asked after all. — πάντα ῥεῖ, Sep 30 '15 at 22:57
Well, if you just can't answer my question it's not because it's poorly asked, it's because you don't know due to lack of knowledge. — John Doe, Sep 30 '15 at 23:00
You had me at _"Suppose I have `int ****ptr;`"_ I mean seriously mate wtf — Lightness Races in Orbit, Sep 30 '15 at 23:45
I'm voting to close this question as off-topic because intellectual exercises require intellect. — Lightness Races in Orbit, Sep 30 '15 at 23:46

score 1 · Accepted Answer · answered Sep 30 '15 at 22:59

1

int ****ptr;
ptr = new int***();
*ptr = new int**();
**ptr = new int*();
***ptr = new int[size_of_arr];

//access (***ptr)[index]

delete[] ***ptr;
delete **ptr;
delete *ptr;
delete ptr;

answered Sep 30 '15 at 22:59

deviantfan

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thank you very much. One last question: parethesis after int*** are used for initialization? What if I don't use them? – John Doe Sep 30 '15 at 23:04
See http://stackoverflow.com/questions/5211090/not-using-parentheses-in-constructor-call-with-new-c – deviantfan Sep 30 '15 at 23:06

score 0 · Answer 2 · answered Sep 30 '15 at 23:08

Note that ***ptr[4] and (***ptr)[4] are not the same thing. The index operator ([4]) takes precedence over the dereference operator (*).

int ****ptr;
ptr = new int***[5];
ptr[4] = new int**;
*ptr[4] = new int*;
**ptr[4] = new int;
***ptr[4] = 77;

delete **ptr[4];
delete *ptr[4];
delete ptr[4];
delete [] ptr;

Allocating array in quadriple pointer

2 Answers2