How does the last line of code dynamically allocate an array of pointers?
int size;
cin >> size;
int** arr = new int* [size];
I am most unclear about the 'int**' part of the code. Can someone break this down?
Thanks!
4 Answers
int a; // a is an int
int *a; // a is a pointer to an int
int **a; // a is a pointer to a pointer to an int
You can make a int* point to an array of int, like this:
int *a = new int[42]; // allocates memory for 42 ints
Exactly the same way, you can make an int** point to an array of int*, like this:
int **a = new int* [42]; // allocates memory for 42 int*
Note that each of the pointers in this array needs to be allocated its own memory, otherwise you just have an array of 42 pointers, none of which are pointing to valid memory.
2 Comments
AKidNamedCoder
I see... so, "int** a;" is pointer which points to an "int pointer", in which that pointer (int** a) is used to dynamically allocate memory for the original "int* a", which is what the ' new int* [size] part of the code does? Am I understanding this correctly?
cigien
Yes, I think you've got it. I'm happy to clarify if you need that though.
Its a double pointer int variable. as I see from code it code. it stores the address of pointer int*
answered Jun 27, 2020 at 17:59
Soumalya Bhattacharya
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Comments
To create a dynamic array the syntax is : data_type * name_of_array = new data_type [size] you can make size as a variable or value as you like .
1 Comment
AKidNamedCoder
Yes, I understand this when referring to a fundamental data type, but I was specifically asking how it works for an array of pointers.
Think of it like this:
To allocate an array you use a type *_var = new type[size]
But you want your type to be pointer to int so: int* *arr = new (int*)[size]
1 Comment
463035818_is_not_an_ai
typo: type vs _type
int* arr = new int[size]allocates an array ofints. It's just that instead ofintyou haveint*.