>Like how struct designated initialisers have to be in order of member declaration due to object lifetime rules or something like that.

It matters in which order sub-objects are initialized - if you have a class with the members A and B, and B takes pointer or reference to A in its constructor and does something with A, A better be already initialized. Sub-objects are initialized in the order of their declaration and having different order of designed initializers would be confusing. In fact, exactly that problem we have in C++ with the list of initializers of base classes and data members in constructors.

You can't easily take the pointer to an other member in a designated initializer. It's still a problem for members that are implicitly initialized by their default member initializer, but that can be sorted out.

I think it is easy enough to be a potential footgun [0]:

  struct A { A(A*);};

  struct B {
    A a1;
    A a2;
  };

  void f()
  {
    B b{.a1 = A(nullptr), .a2 = A(&b.a1) };
  }

[0] https://godbolt.org/z/cGaxzh17T

That's a stretch to call it "easy enough", you are explicitly pointing the gun at your foot. That `b.a1` might not be explicitly UB, but that's quite suspect when b's lifetime didn't start yet. Accessing members through `this` in constructors have some special allowance to not make that UB.

Good point, I should've used direct initialization in this example.

  struct A { 
    int x;
    A(A *a) { if (a) a->x = 42;}
  };

  struct B {
    A a1;
    A a2;
  };

  void f()
  {
    B b{.a1{nullptr}, .a2{&b.a1} };
  }

This code is valid.

Now if I change the struct definition to

  struct B {
    A a2;
    A a1;
  };

it will become UB. Luckily it won't compile because of the difference between the order of declaration and the order of designated initializers.

The alternative way is to always initialize the sub-objects in the order of designated initializers (what do we do if not all initializers are provided?), but this would mean that the order of constructor calls wouldn't match the (reversed) order of destructor calls. Or we would need to select the destructor dynamically based on the way the object was initialized.

https://godbolt.org/z/MPoqEhTvf

My gripe was not the form of initialization of the elements, but forming `b.a1` before `b`'s lifetime has started. It hasn't started before all of the elements are initialized.

But do we need the lifetime of b to be started? Isn't it enough that a1's lifetime is started? Taking of address of a1 happens after that. [0]

Upd:

There is an interesting sentence in [class.cdtor] but I don't think it applies here because B has no constructors:

"For an object with a non-trivial constructor, referring to any non-static member or base class of the object before the constructor begins execution results in undefined behavior."[1]

[0] https://eel.is/c++draft/dcl.init.aggr#7

[1] https://eel.is/c++draft/class.cdtor#1

You still don't need to syntactically require same order initialization, it's an easy job for a compiler to reorder things so that all dependencies work out - every language with order independent declarations have to do that for example.

In a general case you can't do that with separate compilation. [0]

  struct A { A(A*); };

  A* f(struct B *b);

  struct B {
    A a1;
    A a2;
    B(): a1(f(this)), a2(f(this)) {}
  };

  //in a different translation unit
  A* f(B *b)
  {
    return &b->a1; //or a2, we don't know
  }

[0] https://godbolt.org/z/xMb64ssYK

Your code snippet does not use the designated initializer feature that this comment thread is talking about.

Furthermore your code possibly contains undefined behavior depending on the behavior of the constructor of A.

As I stated above, it's the same kind of situation. See my another comment in this thread for an example with designated initializers.

>your code possibly contains undefined behavior

Only if f() returns a pointer to a2 (which is my point). Or did you imply that in the case when f() returns a pointer to a1 and it gets passed to the constructor of a1, provenance matters?

Actually, provenance matters:

"During the construction of an object, if the value of the object or any of its subobjects is accessed through a glvalue that is not obtained, directly or indirectly, from the constructor's this pointer, the value of the object or subobject thus obtained is unspecified." [0]

Reading an unspecified value isn't UB, that's good, but I don't understand why the standard says 'unspecified' because it clearly can be indeterminate if a sub-object hasn't been initialized yet.

[0] https://eel.is/c++draft/class.cdtor#2

You don't need to require same-order initialization, but allowing people to do different orders will be confusing when actions are reordered behind the scenes. Especially imagine if there are dependencies between the objects you're passing in.

  struct A {
    B one;
    C two;
  };
  struct B {
    B() {
      cout << "init B" << endl;
    }
  }
  struct C {
    C() {
      cout << "init C" << endl;
    }
  }

Mixing up the order is confusing:

  A{.two=B(),.one=A()}

since `two` is initialized after `one` despite coming before (the comma operator <expr a>, <expr b> usually means `expr a` happens before `expr b`.

This case is a little contrived, but run the evolution forward: you can have members that depend on each other, or have complex initialization logic of their own. There, debugging the specific order of events is important.

Scala does it by pulling out the keyword arguments to variables so that your example would become

{ val x$1 = B() val x$2 = A() A(.one = x$2, .two = x$1) }

This maintains left-to-right evaluation order while allowing you to pass arguments in any order.

There is probably some dark and forbidden reason why C++ can't do that.

ETA: That's basically what the post does.