Is memory initialization previously guaranteed to persist after placing a new call?

Question

Say I have the following:

struct A { int x; }; //... A* aOriginal = new A(); //value construct aOriginal assert( aOriginal->x == 0 ); A* aSecond = new (aOriginal) A; assert( aSecond->x == 0 ); 

Is the guaranteed second statement guaranteed, although aSecond not value-initialized? Logically, this should, because the memory is not overwritten, but is it specified by the standard?

Answer 1

Not.

When you build a second object in the same storage location, the lifetime of the previous one ends (§3.8 / 1):

[...] The lifetime of an object of type T ends when:

• if T is a class type with a nontrivial destructor (§12.4), the call to the destructor begins, or
• The storage that the object is occupying, reusing, or freed.

When the second object is created, since A has an implicit default constructor, the x member is initialized by default and, therefore, initialization fails (§8.5 / 6):

To initialize an object of type T by default:

• [...]

• otherwise, initialization fails.

And this means that the object has an indefinite value (§5.3.4 / 15):

A new expression that creates an object of type T initializes this object as follows:

• If the new initializer is omitted, the object is initialized by default (§8.5); if initialization fails, the object has an undefined value.

And if you think that the value is not undefined because you previously initialized another object in this repository: the standard discards this possibility, saying also that the properties of the previous object no longer apply after the end of its lifespan (§3.8 / 3):

Properties attributed to objects within the framework of this international standard apply to this object only during its lifetime.