A pointer to an array of pointers

Question

I know I should know this, but it's late and my brain just won't put the pieces together.

This is as straight forward as a question can get:

I have a struct item. I want to create a pointer to an array of pointers to that item type.

Eg.

struct item {
    int data;
    string moreData;
};

I want to have an ArrayPointer that point's to an array. I want that array to contain in each element a pointer to an item.

How do I do this in C++, or more sepcifically where do I need to put how many dereferencing operators? I know how to declare basic (single indirection) pointers and am pretty fluent in their use.

I need information for the following steps if at all possible:

Declaring the ArrayPointer.

Initializing the ArrayPointer with a size s.

Initializing each element of ArrayPointer with new item.

eg:

for(int i = 0; i < s; i++)
    ArrayPointer[i] = // a new item

I feel like as soon as someone posts an answer I'm going to facepalm so hard I break my nose.

Do yourself a factor and use `std::vector` or `std::vector>`. — Captain Obvlious, Nov 23 '14 at 05:30

Vlad from Moscow · Accepted Answer · 2014-11-23T05:48:07.090

1

If I have understood correctly then you need something like this

item **ArrayPointer = new item *[s];

for ( int i = 0; i < s; i++ )
{
    ArrayPointer[i] = new item; { i, "More Data" };
}

Or

item **ArrayPointer = new item *[s];

for ( int i = 0; i < s; i++ )
{
    ArrayPointer[i] = new item;
    ArrayPointer[i]->data = i;
    ArrayPointer[i]->moreData = "More Data";
}

To free the allocated memory you can in reverse order

for ( int i = 0; i < s; i++ )
{
    delete ArrayPointer[i];
}

delete [] ArrayPointer;

Otherewise if s is a constant then you may simply declare an array of pointers. For example

item * ArrayPointer[s];
for ( int i = 0; i < s; i++ )
{
    ArrayPointer[i]->data = i;
    ArrayPointer[i]->moreData = "More Data";
}

edited Nov 23 '14 at 05:48

answered Nov 23 '14 at 05:32

Vlad from Moscow

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Thank's a lot Vlad. The bit I was missing was `new item *[s]` I've never seen that exact statement before. All of my attempts involved `new *items[s]` I just couldn't seem to find that specific syntax anywhere. Thanks again! – user3776749 Nov 23 '14 at 05:39
Sweet. I can also now initiate each element to NULL so I can keep the amount of memory I'm using minimal. – user3776749 Nov 23 '14 at 06:05

Michael Laszlo · Answer 2 · 2014-11-23T05:39:08.690

What you want is an array of struct item *, which are pointers to item structs.

An array of such pointers is a struct item **.

#include <string>
#include <cstdlib>
using namespace std;

struct item {
    int data;
    string moreData;
};

struct item * newItem(int data, string moreData) {
    struct item *result = (struct item *) malloc(sizeof(struct item));
    result->data = data;
    result->moreData = moreData;
    return result;
}

struct item ** array;  // We don't know the size of the array in advance.

int main() {
    int arraySize = 3;  // We get this value from somewhere (user input?).
    array = (struct item **) malloc(3*sizeof(struct item *));
    // Now the array has been allocated. There is space for
    //  arraySize pointers.
    array[0] = newItem(5, "ant");    // Let's make some items. Note that
    array[1] = newItem(90, "bear");  //  newItem() returns a pointer to
    array[2] = newItem(25, "cat");   //  an item.
    return 0;
}

I know how to do that, the problem is that I need the array to be pseudo-dynamic (I need to declare its size once during run-time). The simple version would be: `item *psDynamicArr; psDynamicArr = new item[size];` But in that case the array would contain items, I need it to contain pointers to items. — user3776749, Nov 23 '14 at 05:33
Okay, I've modified the code so that you don't declare its size in advance. — Michael Laszlo, Nov 23 '14 at 05:35

score 0 · Answer 3 · answered Nov 23 '14 at 05:36

0

file.h

struct item {
    int data;
    string moreData;
};

item ** array;

file.cpp

array = new item*[s];

for(int i = 0; i < s; i++)
{
    array[i] = new item;
    array[i]->data = 10;
    array[i]->moreData = "data";
}

answered Nov 23 '14 at 05:36

Ali Mofrad

308
5
21

A pointer to an array of pointers

3 Answers3