CODEWITHSUNDEEP
c++arrayspointersnew-operator
pauliwago
pauliwago

Reputation: 6685

int *array = new int[n]; what is this function actually doing?

I am confused about how to create a dynamic defined array:

 int *array = new int[n];

I have no idea what this is doing. I can tell it's creating a pointer named array that's pointing to a new object/array int? Would someone care to explain?

Upvotes: 30

Views: 174013

Answers (8)

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361682

int *array = new int[n];

It declares a pointer to a dynamic array of type int and size n.

A little more detailed answer: new allocates memory of size equal to sizeof(int) * n bytes and return the memory which is stored by the variable array. Also, since the memory is dynamically allocated using new, you should deallocate it manually by writing (when you don't need anymore, of course):

delete []array;

Otherwise, your program will leak memory of at least sizeof(int) * n bytes (possibly more, depending on the allocation strategy used by the implementation).

Upvotes: 14

Engineero
Engineero

Reputation: 12938

As of C++11, the memory-safe way to do this (still using a similar construction) is with std::unique_ptr:

std::unique_ptr<int[]> array(new int[n]);

This creates a smart pointer to a memory block large enough for n integers that automatically deletes itself when it goes out of scope. This automatic clean-up is important because it avoids the scenario where your code quits early and never reaches your delete [] array; statement.

Another (probably preferred) option would be to use std::vector if you need an array capable of dynamic resizing. This is good when you need an unknown amount of space, but it has some disadvantages (non-constant time to add/delete an element). You could create an array and add elements to it with something like:

std::vector<int> array;
array.push_back(1);  // adds 1 to end of array
array.push_back(2);  // adds 2 to end of array
// array now contains elements [1, 2]

Upvotes: 0

Android M
Android M

Reputation: 31

The new operator is allocating space for a block of n integers and assigning the memory address of that block to the int* variable array.

The general form of new as it applies to one-dimensional arrays appears as follows:

array_var = new Type[desired_size];

Upvotes: 2

LLS
LLS

Reputation: 2228

In C/C++, pointers and arrays are (almost) equivalent. int *a; a[0]; will return *a, and a[1]; will return *(a + 1)

But array can't change the pointer it points to while pointer can.

new int[n] will allocate some spaces for the "array"

Upvotes: 0

aeon
aeon

Reputation: 1069

The statement basically does the following:

  1. Creates a integer array of 'n' elements
  2. Allocates the memory in HEAP memory of the process as you are using new operator to create the pointer
  3. Returns a valid address (if the memory allocation for the required size if available at the point of execution of this statement)

Upvotes: 13

Saurabh Gokhale
Saurabh Gokhale

Reputation: 46415

It allocates that much space according to the value of n and pointer will point to the array i.e the 1st element of array

int *array = new int[n];

Upvotes: 1

DhruvPathak
DhruvPathak

Reputation: 43265

It allocates space on the heap equal to an integer array of size N, and returns a pointer to it, which is assigned to int* type pointer called "array"

Upvotes: 2

ANisus
ANisus

Reputation: 78045

new allocates an amount of memory needed to store the object/array that you request. In this case n numbers of int.

The pointer will then store the address to this block of memory.

But be careful, this allocated block of memory will not be freed until you tell it so by writing

delete [] array;

Upvotes: 39

Related Questions