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

Question

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?

Answer 1

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;

Answer 2

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've to 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).

Answer 3

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)

Answer 4

它在堆上分配的空间等于大小为N的整数数组，并返回指向它的指针，该指针被分配给称为“数组”的int *类型指针

Answer 5

它根据n的值分配那么多空间，指针将指向数组，即数组的第一个元素

int *array = new int[n];

Answer 6

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]

Answer 7

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"

Answer 8

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];