https://gcc.gnu.org/bugzilla/show_bug.cgi?id=63526
--- Comment #5 from Dávid Éles <eles.david.88 at gmail dot com> --- (In reply to Daniel Krügler from comment #3) > (In reply to Dávid Éles from comment #2) > > I uses the default mechanism to initialization of members. > > As far as I know the C++ standard says (8.5/5): > > To default-initialize an object of type T means: > > * If T is a non-POD class type (clause 9), the default constructor for T is > > called (and the initialization is ill-formed if T has no accessible default > > constructor). > > [Based on your quotes and your compiler settings you seem to quote the > C++98/03 standard. This is where my response is referred to as well] > > In your example an object of type Foo<double> is default-initialized. Class > Foo<double> is a non-POD class type, and therefore exactly this bullet is > entered. The result is what the wording says: the default constructor for T > (that is Foo<double> in this example is called. The semantics of calling the > default-constructor of a class that has no member-initializer provided (such > as in this case) is specified in [class.base.init] p5: > > "If a given nonstatic data member or base class is not named by a > mem-initializer-id (including the case where there is no > mem-initializer-list because the constructor has no ctor-initializer), then > — If the entity is a nonstatic data member of (possibly cv-qualified) class > type (or array thereof) or a base class, and the entity class is a non-POD > class [..] > — Otherwise, the entity is not initialized. [..] > > The first bullet here does not apply, because the member is of type double > and thus does not match a class type. The second bullet therefore > unconditionally applied and says that the member is not initialized at all. > > > * If T is an array type, each element is default-initialized. > > This bullet does not apply > > > * Otherwise, the object is zero-initialized. > > This bullet does not apply. > > > In case of c++ it should be zero initialized if it is the member of a > > class/struct. > > No, you are incorrectly interpreting the Standard. > > > As far as I know I have to force to not doing zero initialization something > > like that Foo* f = new Foo; > > That has essentially the same initialization semantics as your example code. You are right, I missed it, thanks for your quick answer. Sorry to waste your time.