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Do the parentheses after the type name make a difference with new?

If 'Test' is an ordinary class, is there any difference between:

Test* test = new Test; 

and

Test* test = new Test(); 
like image 292
David Read Avatar asked Mar 06 '09 19:03

David Read


2 Answers

Let's get pedantic, because there are differences that can actually affect your code's behavior. Much of the following is taken from comments made to an "Old New Thing" article.

Sometimes the memory returned by the new operator will be initialized, and sometimes it won't depending on whether the type you're newing up is a POD (plain old data), or if it's a class that contains POD members and is using a compiler-generated default constructor.

  • In C++1998 there are 2 types of initialization: zero and default
  • In C++2003 a 3rd type of initialization, value initialization was added.

Assume:

struct A { int m; }; // POD struct B { ~B(); int m; }; // non-POD, compiler generated default ctor struct C { C() : m() {}; ~C(); int m; }; // non-POD, default-initialising m 

In a C++98 compiler, the following should occur:

  • new A - indeterminate value
  • new A() - zero-initialize

  • new B - default construct (B::m is uninitialized)

  • new B() - default construct (B::m is uninitialized)

  • new C - default construct (C::m is zero-initialized)

  • new C() - default construct (C::m is zero-initialized)

In a C++03 conformant compiler, things should work like so:

  • new A - indeterminate value
  • new A() - value-initialize A, which is zero-initialization since it's a POD.

  • new B - default-initializes (leaves B::m uninitialized)

  • new B() - value-initializes B which zero-initializes all fields since its default ctor is compiler generated as opposed to user-defined.

  • new C - default-initializes C, which calls the default ctor.

  • new C() - value-initializes C, which calls the default ctor.

So in all versions of C++ there's a difference between new A and new A() because A is a POD.

And there's a difference in behavior between C++98 and C++03 for the case new B().

This is one of the dusty corners of C++ that can drive you crazy. When constructing an object, sometimes you want/need the parens, sometimes you absolutely cannot have them, and sometimes it doesn't matter.

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Michael Burr Avatar answered Sep 19 '22 23:09

Michael Burr


new Thing(); is explicit that you want a constructor called whereas new Thing; is taken to imply you don't mind if the constructor isn't called.

If used on a struct/class with a user-defined constructor, there is no difference. If called on a trivial struct/class (e.g. struct Thing { int i; };) then new Thing; is like malloc(sizeof(Thing)); whereas new Thing(); is like calloc(sizeof(Thing)); - it gets zero initialized.

The gotcha lies in-between:

struct Thingy {   ~Thingy(); // No-longer a trivial class   virtual WaxOn();   int i; }; 

The behavior of new Thingy; vs new Thingy(); in this case changed between C++98 and C++2003. See Michael Burr's explanation for how and why.

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kfsone Avatar answered Sep 19 '22 23:09

kfsone