Default initialization in C++

Question

I was asking myself something this morning, and I can\'t find the words to properly \"google\" for it:

Lets say I have:

struct Foo
{
  int bar;
};

s         


        

Answer 1

Since bar is a built-in type its default initialization will be undefined for Foo1 and Foo2. If it would have been a custom type, then the default constructor would have been called, but here it's not the case.

Lesson: always initialize your variables.

Answer 2

Case 3 is the proper way, with a member initialization list.

None of the first two will be properly initialized, since you don't give them an initial value (exactly like a variable only defined is not initialized).

Answer 3

For pod-types, the default initialization is zero-initialization.

Therefore:

Foo() : b() {} is the same as Foo() : b(0) {}

I can't find the respective part of the C++ standard, but if you skip the initializer completely, then POD types shouldn't be default-initialized (unlike non-POD types, which are).

Therefore in your case, only the third example is correctly initialized.

Answer 4

Only foo3 will be in all contexts. foo2 and foo will be if they are of static duration. Note that objects of type Foo may be zero initialized in other contexts:

Foo* foo = new Foo(); // will initialize bar to 0
Foo* foox = new Foo; // will not initialize bar to 0

while Foo2 will not:

Foo2* foo = new Foo2(); // will not initialize bar to 0
Foo2* foox = new Foo2; // will not initialize bar to 0

that area is tricky, the wording as changed between C++98 and C++03 and, IIRC, again with C++0X, so I'd not depend on it.

With

struct Foo4
{
   int bar;
   Foo4() : bar() {}
};

bar will always be initialized as well.