CODEWITHSUNDEEP
c++arrayspointers
Viliami
Viliami

Reputation: 628

Difference between int * array[60] and int * array = new int(60);

int * array[60]; //creates an array of 60 pointers to an int
int * array = new int(60); //same thing?

Do these both result in the same type of array? e.g an array of pointers to integers

I know that the first one uninitialized, and the second one is initialized, but I am unsure of what exactly the second one creates.

Upvotes: 2

Views: 1619

Answers (5)

user6846842
user6846842

Reputation:

One of them creates an array, the other doesn't.

int * array[60]; // Creates an array of 60 integer pointers

Upvotes: 0

songyuanyao
songyuanyao

Reputation: 172924

int * array = new int(60); //same thing?

No, they're not the same thing. array is just a pointer here, and then point to an int with initialized value 60.

If you meant int * array = new int[60];, array will point to an array of 60 ints, they're still not the same thing.

Note that just as the declaration, int* array means it is a pointer, while int* array[60] means it is an array (of 60 pointers). (Array might decay to pointer, i.e. int** for int* array[60], it's not same as int*.)

Upvotes: 11

EFenix
EFenix

Reputation: 831

To help understand the difference, take into account that the first line creates a 60 pointers-block in memory (in the stack if you are inside main, for example), but the second one only a pointer block.

Both are completely different types. For example, try the following:

array++;

In the first line, it does not work. In the second one, it's ok. If you try:

array[0]++;

In the first line you change the first pointer, in the second one you add 1 to the integer (change to 61).

Upvotes: -1

txid
txid

Reputation: 66

Here are two pictures that illustrate the differences between 

int * array[5];

and 

int * array = new int(5);

To create a pointer int array use int * array = new int[5];

code,

debug view

Upvotes: 0

asimes
asimes

Reputation: 5894

Perhaps you do not realize that the second case is not an array, the following program prints 60:

#include <iostream>

int main() {
    int* foo = new int(60);
    std::cout << *foo << '\n';
    return 0;
}

Upvotes: 2

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