C++ class initialization with and without parentheses

Question

What is the difference between the following 2 two initializations?

class Pod {
public:
    int a, b;
};

Pod *p1 = new Pod;
Pod *p2 = new Pod();

As JamesMcLaughlin said, no difference, furthermore that's not limited to PODs. — MFH, Mar 17 '13 at 02:16
@DavidRodríguez-dribeas Yep, looks like I was wrong! Thanks. — James M, Mar 17 '13 at 02:21
This previous thread explains zero, default and value-initialized and has examples http://stackoverflow.com/questions/1613341/what-do-the-following-phrases-mean-in-c-zero-default-and-value-initializat — Shafik Yaghmour, Mar 17 '13 at 02:41
This is similar to the "most vexing parse", http://stackoverflow.com/questions/180172/default-constructor-with-empty-brackets see http://ideone.com/ompWis — QuentinUK, Mar 17 '13 at 03:34

score 11 · Accepted Answer · answered Mar 17 '13 at 02:16

11

In the first case the object is left uninitialized, while in the second case the object is guaranteed to be value-initialized, which in this case as the type is POD it means zero-initialized

answered Mar 17 '13 at 02:16

David Rodríguez - dribeas

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Though the difference is essentially nonexistent in the case of `int`, in the first case the members are default initialized rather than uninitialized. – Jerry Coffin Mar 17 '13 at 02:22
@JerryCoffin: Yes I should have been clearer, that part is also due to the fact that it is a POD, for which *default-initialization* means that the object is *left uninitialized* – David Rodríguez - dribeas Mar 17 '13 at 02:39
1

@DavidRodríguez-dribeas: ...but in C++11, it's not restricted to POD types either (IIRC, it should apply to the somewhat broader category of all trivially copyable types). – Jerry Coffin Mar 17 '13 at 02:57
@JerryCoffin: *trivially copyable types*? Not sure how that would matter here (there are no copies anywhere) and I cannot find a reference in n3777 that supports that. For *value-initialization* the distinction is between a class type with/without user provided default constructor. For *default-initialization* of a class type it implies calling the default constructor, which if defined implicitly will recursively cause *default-initialization* of the members... at the end members of fundamental types are left uninitialized – David Rodríguez - dribeas Mar 17 '13 at 03:16
1

@DavidRodríguez-dribeas: Copying isn't really relevant, but being trivially copyable is a looser requirement than being POD, and you don't have to meet the extra requirements to be POD for default initialization to mean leaving uninitialized. – Jerry Coffin Mar 17 '13 at 03:43

C++ class initialization with and without parentheses

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