Is a local scoped variable initialized to an undetermined value, or un-initialized?

Question

Pedantically speaking, is x initialized in the following code or not?

int main()
{
    int x;
}

There are some paragraphs about it in 8.5 Initializers [dcl.init] (for C++11) but not backed by any examples.

It is default-initialized. Which means that no initialization is performed (for `int`). — dyp, Sep 23 '14 at 15:47
Thus it stays indeterminate, so reading it (as it is auto-storage-class) is UB. — Deduplicator, Sep 23 '14 at 15:50

dyp · Answer 1 · 2014-09-23T15:54:32.223

5

It is formally default-initialized, which means for ints, that no initialization is performed.

[dcl.init]/12 (N3797)

If no initializer is specified for an object, the object is default-initialized; if no initialization is performed, an object with automatic or dynamic storage duration has indeterminate value

[dcl.init]/7

To default-initialize an object of type T means:

if T is a (possibly cv-qualified) class type, the default constructor for T is called [...];

if T is an array type, each element is default-initialized;

otherwise, no initialization is performed.

edited Sep 23 '14 at 15:54

answered Sep 23 '14 at 15:49

dyp

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So, it is initialized, but that action consists of "no initialization is performed". Either that, or it is default-initialized, but default-initialized is not a kind of initialized in this case. Or, the term "initialized" has no formal meaning in the standard, while "default-initialized" does as does "initializaion is performed". Gah. – Yakk - Adam Nevraumont Sep 23 '14 at 15:50
@Yakk Exactly. Reminds me of complete types that cannot be completed. – dyp Sep 23 '14 at 15:53
@Yakk: It stays indeterminate is thus factually correct and the standard conformant wording for its state. So, using it if it's auto-storage-class is UB according to 4.1 lvalue-to-rvalue-conversion [conv.lval]. – Deduplicator Sep 23 '14 at 15:53
@Deduplicator thankfully [in C++14](http://stackoverflow.com/q/23415661/1708801) we don't need to rely on the l-to-r conversion to know this undefined behavior. – Shafik Yaghmour Sep 23 '14 at 16:05

score 3 · Answer 2 · edited Jun 20 '20 at 09:12

No, it isn't. According to standard, x is default-initialized ([dcl.init]/6):

To default-initialize an object of type T means:

— if T is a (possibly cv-qualified) class type [...]

— if T is an array type [...]

— otherwise, no initialization was performed.

x is therefore uninitialized since no initialization is performed.
Hence the object has indeterminate value ([dcl.init]/11):

If no initializer is specified for an object, the object is default-initialized; if no initialization is performed, an object with automatic or dynamic storage duration has indeterminate value.

Moreover, if we were to access it's stored, indeterminate value - in other words, perform an lvalue-to-rvalue conversion on it - we would induce undefined behavior ([conv.lval]/1]):

If the object to which the glvalue refers is [..], or if the object is uninitialized, a program that necessitates this conversion has undefined behavior.

score 2 · Answer 3 · answered Sep 23 '14 at 16:04

2

The way I understand it is that the place in memory for the variable x is reserved, but not set to a value (un-initialized). Because it is un-initialized, any old values there will be considered as 'garbage' int.

answered Sep 23 '14 at 16:04

Abdulrahman Alhadhrami

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Is a local scoped variable initialized to an undetermined value, or un-initialized?

3 Answers3