What does the () at the end of dynamic array allocate mean?

Question

I have see some example like below in a different question which I have not seen before.

new int[m_size]();
               ^^

I have seen and used the version new int[m_size] all the time but not one with the () at the end.

http://stackoverflow.com/questions/620137/do-the-parentheses-after-the-type-name-make-a-difference-with-new — Lightness Races in Orbit, Apr 05 '11 at 12:35

score 12 · Accepted Answer · edited Jun 20 '20 at 09:12

12

Two words : Value Initialization

new int[m_size](); array elements would be zero-initialized by writing () because () implies value initialization.¹ (zero initialization for a primitive type)

_{1: An object whose initializer is an empty set of parentheses, i.e., (), shall be value-initialized. ( $8.5/7 )}

edited Jun 20 '20 at 09:12

Community

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answered Apr 05 '11 at 12:33

Prasoon Saurav

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ah! So which type does it value initialise? Is it the pointer type of the whole expression or the individual members of the array? – dubnde Apr 05 '11 at 12:37
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@MeThinks : Check out `8.5/5` (Value Initialization). For a primitive type **Value initialization** => **Zero Initialization**. In this case the all the elements of the array would be initialized to `0` – Prasoon Saurav Apr 05 '11 at 12:39
ok. Got it in the standard "if T is an array type, then each element is value-initialized". Thanks for the prompt response. First time I've come across it with respect to arrays. – dubnde Apr 05 '11 at 12:41

score 2 · Answer 2 · answered Apr 05 '11 at 12:39

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it means all the elements will be zero initialized,similar to calloc(o,sizeof(int)) where with this calloc ,ur initializing a single integer on heap with 0

answered Apr 05 '11 at 12:39

Vijay

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What does the () at the end of dynamic array allocate mean?

2 Answers2