Is a float member guaranteed to be zero initialized with {} syntax?

Question

In C++17, consider a case where S is a struct with a deleted default constructor and a float member, when S is initialized with empty braces, is the float member guaranteed by the standard to be zero-initialized?

struct A {
  int x{};
};

struct S
{
  S() = delete;
 
  A a;
  float b;
};

int main()
{
  auto s = S{}; // Is s.b guaranteed to be zero?
}

In my opinion, cppreference.com is not clear, saying both that:

If the number of initializer clauses is less than the number of members and basesor initializer list is completely empty, the remaining members and bases (since C++17) are initialized by their default member initializers, if provided in the class definition, and otherwise (since C++14) copy-initialized from empty lists, in accordance with the usual list-initialization rules (which performs value-initialization for non-class types and non-aggregate classes with default constructors, and aggregate initialization for aggregates). If a member of a reference type is one of these remaining members, the program is ill-formed.

(from here), which implies that b is guaranteed to be zero

In all cases, if the empty pair of braces {} is used and T is an aggregate type, aggregate-initialization is performed instead of value-initialization.

(from here)

which implies that b is not guaranteed to be zero.

There is also a discussion that seems to imply that while not guaranteed, all known compiler zero-initialize anyway:

The standard specifies that zero-initialization is not performed when the class has a user-provided or deleted default constructor, even if that default constructor is not selected by overload resolution. All known compilers performs additional zero-initialization if a non-deleted defaulted default constructor is selected.

Related to Why does aggregate initialization not work anymore since C++20 if a constructor is explicitly defaulted or deleted?

I don't know... Is the delete not supposed to prevent default construction? Or is it still the case that a stupid C++ rule allows side-stepping the deletion? — Johannes Schaub - litb, Oct 18 '21 at 08:39
The provided example compiles. It does not call the default construtor but does aggregate initialization — fouronnes, Oct 18 '21 at 08:42
That's great news! My question is explicitly for C++17 though. — fouronnes, Oct 18 '21 at 08:43
@StoryTeller-UnslanderMonica Do you have a link about the C++20 fix? — fouronnes, Oct 18 '21 at 08:46
This is the paper that was voted into C++20 https://wg21.link/P1008R1 — StoryTeller - Unslander Monica, Oct 18 '21 at 08:48
The second quote comes from the value-initialization page, which says that your structure will be aggregate-initialized. So that means the first quote is the one that details how the structure will be initialized. And that quote says that the members will be [list-initialized](https://en.cppreference.com/w/cpp/language/list_initialization). And since no list is provided, the members will be individually value-initialized (or aggregate-initialized, depending on type). So `b` will be value-initialized and zero. — Some programmer dude, Oct 18 '21 at 08:55
@StoryTeller-UnslanderMonica thanks for the link. I've discovered the `explicit` trick and answered myself thanks to you. Technically I'm still looking for the answer about whether b will be zero initialized or not, but practically the explicit trick solves my real problem! — fouronnes, Oct 18 '21 at 08:56
Happy to help. I feel I should point out that this obscure quirk did have one upside. It allowed you to write aggregates that did not permit indeterminate values. Consider https://wandbox.org/permlink/aq1t3bMV9ORogvMy - And notice how only the default (and problematic) initialization is disabled. I still agree that the rule was too arcane and weird to stay, but it wasn't all bad. — StoryTeller - Unslander Monica, Oct 18 '21 at 08:59
Oh I see. That's actually pretty nice. In C++20 you'd need to manually provide the trivial constructor that forwards the arguments from the initializer list to mimick aggretate initialization. I see. — fouronnes, Oct 18 '21 at 09:14
@fouronnes [The fickle aggregate](https://dfrib.github.io/the-fickle-aggregate/). — dfrib, Oct 18 '21 at 12:31

fouronnes · Answer 1 · 2021-10-18T20:55:20.580

6

This is a quirk of C++ that is fixed in C++20. In the meantime you can add explicit to the deleted default constructor to force the struct to become non-aggregate, and make your code a guaranteed compile error:

struct A {
  int x{};
};

struct S
{
  explicit S() = delete;
 
  const A a;
  const float b;
};

int main()
{
  auto s = S{}; // error: call to deleted constructor of 'S'
}

edited Oct 18 '21 at 20:55

answered Oct 18 '21 at 08:55

fouronnes

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score 5 · Accepted Answer · answered Oct 18 '21 at 13:29

Because S is an aggregate, S{} will perform aggregate initialization. The rule in the standard about how members are initialized when there are no initializers in the list is basically what you cited:

If the element has a default member initializer ([class.mem]), the element is initialized from that initializer.

Otherwise, if the element is not a reference, the element is copy-initialized from an empty initializer list ([dcl.init.list]).

So for b, that's the equivalent of float b = {};. Per the rules of list initialization, we have to get all the way down to 3.10:

Otherwise, if the initializer list has no elements, the object is value-initialized.

And value initialization will initialize a float to 0.

Is a float member guaranteed to be zero initialized with {} syntax?

2 Answers2