Does new[] call default constructor in C++?

Question

When I use new[] to create an array of my classes:

int count = 10;
A *arr = new A[count];

I see that it calls a default constructor of A count times. As a result arr has count initialized objects of type A. But if I use the same thing to construct an int array:

int *arr2 = new int[count];

it is not initialized. All values are something like -842150451 though default constructor of int assignes its value to 0.

Why is there so different behavior? Does a default constructor not called only for built-in types?

Solution 1

See the accepted answer to a very similar question. When you use new[] each element is initialized by the default constructor except when the type is a built-in type. Built-in types are left unitialized by default.

To have built-in type array default-initialized use

new int[size]();

Solution 2

Built-in types don't have a default constructor even though they can in some cases receive a default value.

But in your case, new just allocates enough space in memory to store count int objects, ie. it allocates sizeof<int>*count.

Comments

[James Curran]

@rubber boots: int i (); does not initialize a varaible named i. It declares a function i returning an int. You may have meant int i = int();

[fredoverflow]

@James: In C++0x you can finally say what you mean: int x{}; :)

[rubber boots]

@James, oops - wtf did I write? Thanks for clearing this up. Wrong deduction from int *i = new int();. @Cerdic, Sorry for Posting BS.

[aardvarkk]

Wow, I didn't actually know you could add the braces to default initialize built-in types. I've been working with C++ for years. That's either exciting or very sad!