Does the default constructor initialize built-in t

Question

Does the default constructor (created by the compiler) initialize built-in-types?

Answer 1

For all practical purposes - no.

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However for implementations that are technically compliant with the C++ standard, the answer is that it depends whether the object is POD or not and on how you initialize it. According to the C++ standard:

MyNonPodClass instance1;//built in members will not be initialized
MyPodClass instance2;//built in members will be not be initialized
MyPodClass* instance3 = new MyPodClass;//built in members will not be initialized
MyPodClass* instance3 = new MyPodClass() ;//built in members will be zero initialized

However, in the real world, this isn't well supported so don't use it.

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The relevant parts of the standard are section 8.5.5 and 8.5.7

Answer 2

As previous speakers have stated - no, they are not initialized.

This is actually a source for really strange errors as modern OSs tend to fill newly allocated memory regions with zeroes. If you expect that, it might work the first time. However, as your application keeps running, delete-ing and new-ing objects, you will sooner or later end up in a situation where you expect zeroes but a non-zero leftover from an earlier object sits.

So, why is this then, isn't all new-ed data newly allocated? Yes, but not always from the OS. The OS tends to work with larger chunks of memory (e.g. 4MB at a time) so all the tiny one-word-here-three-bytes-there-allocations and deallocations are handled in uyserspace, and thus not zeroed out.

PS. I wrote "tend to", i.e. you can't even rely on success the first time...

Answer 3

Technically it does initialize them -- by using their default constructor, which incidentally does nothing but allocate the memory for them.

If what you wanted to know is whether or not they are set to something sane like 0 for ints, then the answer is "no".

Answer 4

Implicitly defined (by the compiler) default constructor of a class does not initialize members of built-in types.

However, you have to keep in mind that in some cases the initialization of a instance of the class can be performed by other means. Not by default constructor, nor by constructor at all.

For example, there's a widespread incorrect belief that for class C the syntax C() always invokes default constructor. In reality though, the syntax C() performs so called value-initialization of the class instance. It will only invoke the default constructor if it is user-declared. (That's in C++03. In C++98 - only if the class is non-POD). If the class has no user-declared constructor, then the C() will not call the compiler-provided default constructor, but rather will perform a special kind of initialization that does not involve the constructor of C at all. Instead, it will directly value-initialize every member of the class. For built-in types it results in zero-initialization.

For example, if your class has no user-declared constructor

class C { 
  int x;
};

then the compiler will implicitly provide one. The compiler-provided constructor will do nothing, meaning that it will not initialize C::x

C c; // Compiler-provided default constructor is used
// Here `c.x` contains garbage

Nevertheless, the following initializations will zero-initialize x because they use the explicit () initializer

C c = C(); // Does not use default constructor for `C()` part
           // Uses value-initialization feature instead
assert(c.x == 0);

C *pc = new C(); // Does not use default constructor for `C()` part
                 // Uses value-initialization feature instead
assert(pc->x == 0);

The behavior of () initializer is different in some respects between C++98 and C++03, but not in this case. For the above class C it will be the same: () initializer performs zero initialization of C::x.

Another example of initialization that is performed without involving constructor is, of course, aggregate initialization

C c = {}; // Does not use any `C` constructors at all. Same as C c{}; in C++11.
assert(c.x == 0);

C d{}; // C++11 style aggregate initialization.
assert(d.x == 0);

Answer 5

I'm not quite certain what you mean, but:

struct A { int x; };

int a; // a is initialized to 0
A b;   // b.x is initialized to 0

int main() {
    int c;         // c is not initialized
    int d = int(); // d is initialized to 0

    A e;           // e.x is not initialized
    A f = A();     // f.x is initialized to 0
}

In each case where I say "not initialized" - you might find that your compiler gives it a consistent value, but the standard doesn't require it.

A lot of hand-waving gets thrown around, including by me, about how built-in types "in effect" have a default constructor. Actually default initialization and value initialization are defined terms in the standard, which personally I have to look up every time. Only classes are defined in the standard to have an implicit default constructor.

Answer 6

As per the standard, it doesn't unless you explicitly initialize in initializer list