Freeing memory with Pointer Arithmetic

Question

C++ newb here. I'm trying to write my own implementation of an array using only pointers, and I've hit a wall I don't know how to get over.

My constructor throws this error

array.cpp:40:35: error: invalid conversion from ‘int*’ to ‘int’ [-fpermissive]

When my array initializes I want it to free up all the spaces in the array for ints.

Array::Array(int theSize){
size = theSize;
int *arrayPointer = new int; 
int index = 0;
while(theSize > index){
  *(arrayPointer + index) = new int; //This is the trouble line.
  ++index;
 }
}

What am I doing wrong stackoverflow?

Well, for starters you're attempting to allocate each element individually, when there's a built-in `new int[size]` syntax for array allocation ;-) — Cameron, Aug 29 '12 at 21:33
Using `new int[size]` is just as bad (possibly worse). For a container, you normally want to use `buffer = operator new(size);` to allocate raw storage, followed by placement new to create objects in that space: `new(buffer+pos) int(value);` This matters little with `int`, but much more with other types. — Jerry Coffin, Aug 30 '12 at 00:01
Since this comes up so often, and nearly always produces a slew answers I think are mostly wrong, I wrote up a [blog post](http://coderscentral.blogspot.com/2012/08/c-dynamic-arrays.html) about what I consider a better way to handle this. — Jerry Coffin, Aug 30 '12 at 03:47

score 1 · Accepted Answer · edited May 23 '17 at 10:24

1

arrayPointer points to a single int, it does not point to an array of int*, which this line would require:

*(arrayPointer + index) = new int;

but the type of *(arrayPointer + index) is an int, hence the compiler error.

To allocate an array of int:

int* arrayPointer = new int[size];

If this is intended to initialise a member variable then:

arrayPointer = new int[size];

otherwise arrayPointer would be local to the constructor. As the class now has a dynamically allocated member you need to either implement both copy constructor and assignment operator or prevent copying (see What is The Rule of Three?). Remember to delete[] arrayPointer in the destructor.

Just mentioning std::vector<int>, even though this is a learning exercise.

edited May 23 '17 at 10:24

Community

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answered Aug 29 '12 at 21:33

hmjd

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I tried new int[size] and the program hurled. Is there an include I need to use? – Breedly Aug 29 '12 at 21:39
Did you redeclare `arrayPointer` in the constructor (see updated answer)? – hmjd Aug 29 '12 at 21:40
Still hurling, forgot to post the error. `array.cpp:(.text+0x1c): undefined reference to operator new[] (unsigned int)'` Thanks for the link, I'll read more into it. – Breedly Aug 29 '12 at 21:50
@Breedly, what are compiler are using? GNU? If so, use `g++` and not `gcc`. – hmjd Aug 29 '12 at 21:52
Hah, problem solved, only to add more problems out of the scope of this question! Thanks so much, you were a huge help. Feel good about yourself for a few days. – Breedly Aug 29 '12 at 21:58

score 1 · Answer 2 · edited Aug 29 '12 at 23:16

1

Do as follows:

#include <cstddef>

template <typename T>
class Array
{

public:

    T* const arrayPointer; // arrayPointer can't be reallocated
    const size_t size; // size can't change

    Array(const int theSize) : arrayPointer(new T[theSize]),
                               size(theSize) {}

    ~Array() {
        delete[] arrayPointer;
    }

private:

    Array(const Array& other) {} // forbid copy

    Array& operator= (const Array& other) {} // forbid assignment

} ;

Why to use template <typename T>? So you can have an array of any type.
Why to use new T[ theSize ] ? So you can allocate theSize elements at the same time.
Why to use : arrayPointer( new T[ theSize ]) ? So in case allocation fails (due to big theSize) the object fails without initialization. It is callled RAII.
Why to use delete [] arrayPointer ? Because you used new[] and you have to deallocate the whole array.
Why those const's? To avoid anyone changing the size of the array and making the fields inconsistent.
What are these private methods? They are forbiding copies, so no one can make array1 = array2; delete array2;, what would deallocate the arrayPointer of array1.

Usage (it will allocate array of 10 int:)

Array< int > arr( 10 ) ;

Access:

arr.arrayPointer[ 0 ] = 5 ;

Note - you can access arrayPointer at the cells of range 0..9. You can add operator[] to your class in order to avoid using arrayPointer and to use arr[ 0 ].

edited Aug 29 '12 at 23:16

Alex Oliveira

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answered Aug 29 '12 at 21:41

Grzegorz

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This is also a great answer, but it goes a little out of scope. You should feel good about yourself too for the next couple of days! – Breedly Aug 29 '12 at 21:59
@Breedly: I have just started feeling good about myself ;) Seriously, since you are a newbee to C++ then why not start the right way? And since you want to implement your ownr array, I think it is the best and safest way too. Have fun with C++. – Grzegorz Aug 29 '12 at 22:17
@Grzegorz Nice things to teach, but maybe too early. Now that you have shown that he needs a Destructor, you probably need to talk about the Rule of Three, otherwise it will not be "the best and safest way". – Alex Oliveira Aug 29 '12 at 22:42
@AlexOliveira: There is never end to how much one can learn ;) Yes, there is so much into it... pure C++ is really cool. – Grzegorz Aug 29 '12 at 22:52
@Grzegorz: Agree! I tried to edit you code to make it adhere to the Rule of Three without complicating it too much. – Alex Oliveira Aug 29 '12 at 23:15
@AlexOliveira: Thanks for the effort. Approved. – Grzegorz Aug 29 '12 at 23:18

score 0 · Answer 3 · answered Aug 29 '12 at 21:36

The line you mentioned is trying to set a int* into a int var => Dereferencing a int* provide a int :

*(arrayPointer + index) // is a int

Anyway you are moving in memory (and dereferencing memory) you haven't reserved. So you can access a protected memory zone by doing this instruction.

Replace all your constructor with:

Array::Array(int theSize)
{
    size = theSize;
    arrayPointer = new int[theSize]; // Replace your local var with a member one. Else you will lose your array ;)
}

Freeing memory with Pointer Arithmetic

3 Answers3