Is there a benefit to declaring the default constructor as default?

Question

In C++17, does it make any difference to declare the default constructor as default in an aggregate, as in:

struct A {
   A() = default;
   int v;
};

vs relying on it being implicitly defined, as in:

struct A {
   int v;
};

I understand that in C++20, declaring the constructor causes A to not be an aggregate. As I am switching our build flag to C++20, I'd like to just remove the constructor declaration so the structs remain aggregates, but I'm wondering is this has any side effect?

Well, if you have another one, it saves you typing (or even not). — πάντα ῥεῖ, Apr 21 '23 at 20:15
`A() = default;` is a user declared constructor, sometimes that matters — 463035818_is_not_an_ai, Apr 21 '23 at 20:16
In C++17 there is no difference but in C++20 there is. The real question you have to ask yourself is do you want your type to be considered an aggregate or not. — NathanOliver, Apr 21 '23 at 20:18
https://stackoverflow.com/questions/4178175/what-are-aggregates-and-pods-and-how-why-are-they-special — 463035818_is_not_an_ai, Apr 21 '23 at 20:22
You can use `type_traits` and `std::is_aggregate`, as [shown here](https://godbolt.org/z/bcnEqKhW5). — PaulMcKenzie, Apr 21 '23 at 20:24

score 0 · Answer 1 · answered Apr 21 '23 at 20:32

0

does it make any difference to declare the default constructor as default in an aggregate

No differences in this case. Regardless whether you explicitly writes a default constructor =default, the compiler generates the same thing, and your variable v is initialized to 0 anyway.

answered Apr 21 '23 at 20:32

artm

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`v` will not be initialized to zero by the defaulted default constructor. – NathanOliver Apr 21 '23 at 20:44
ok so it's "default" value is initialized when we define it eg `struct A a1;' ? – artm Apr 21 '23 at 20:59
@artm There is no "_"default" value_". The constructor doesn't change the value of the `int` member at all (whether implicitly-defined or explicitly defaulted) and leaves it with an indeterminate value (or zero value if the object was zero-initialized beforehand). – user17732522 Apr 21 '23 at 21:03
For built in types like `int`, default initialization is no initialization, the `int` just has some garbage value. `A foo;` leaves `v` with an indeterminate value. `A foo{}` (or if you could write `A foo()`) value initializes `foo` and that will zero initialize `v`. – NathanOliver Apr 21 '23 at 21:04
@NathanOliver so you mean in this case `v` is never initialised to zero - regardless of whether the default ctor is there or not? there is this related answer talking about "value-initialised" - not sure if it's exactly the same here: https://stackoverflow.com/a/3803288 – artm Apr 21 '23 at 22:15
If you do `A foo;` then yes, `v` will not be initialized as you are doing default initialization. – NathanOliver Apr 21 '23 at 22:22
@NathanOliver and `A foo();` then `v` will be initialized - correct ? – artm Apr 21 '23 at 22:23
1

Yes, except that you can't do `A foo()` as that declares a function. You have to do `A foo{};` or `A foo = A();` – NathanOliver Apr 21 '23 at 22:29

Is there a benefit to declaring the default constructor as default?

1 Answers1