Aggregate / designated initialization of c++ struct: Refer directly to another field

Question

When using aggregate / designated initialization of a struct it is possible to refer to another field like this:

#include <stdio.h>


int main()
{
  struct
  {
    int a;
    int b;
  }
  s = 
  {
    .a = 3,
    .b = s.a + 1,
  };

  return 0;
}

We use s.a in the initialization of s.b. However, we need to refer to s.a through s. Is it possible to refer directly to s.a as just .a, or similar? This would for instance make it possible to use the same syntax when initializing arrays of structs like here:

int main()
{
  struct
  {
    int a;
    int b;
  }
  s[] = 
  {
    {
      .a = 3,
      .b = s[0].a + 1,
    },
    {
      .a = 5,
      .b = s[1].a - 1,
    }
  };

  return 0;
}

Here we could maybe write something like .b = .a - 1 instead of .b = s[1].a - 1.

Even if you could refer directly to a member in the initializing expression, how would that help in the array case? Given `.b = .a - 1` how would you distinguish between `.b = s[0].a - 1` and `.b = s[1].a - 1`? — cigien, Nov 12 '20 at 19:05
@cigien They would be distinguished through the scope, like if `.a` was a normal variable. — Sara Haugen, Nov 12 '20 at 19:10
Ok, let's say that worked, would `.a` refer to `s[0].a` or `s[1].a`? Both are in the same scope. — cigien, Nov 12 '20 at 19:11
@cigien If the brace initialiers for each struct were treated as a scope, then they would be in different scopes. Of course this is probably a theoretical discussion since it doesn't seem to work like that. — Sara Haugen, Nov 12 '20 at 19:17
Indeed, it doesn't work like that, but there's nothing wrong with theory :) I see now what you want the code to do, I misunderstood that. And yes, that's a reasonable thing to expect the code to do, except for the rules not allowing it :p — cigien, Nov 12 '20 at 19:24

dfrib · Accepted Answer · 2020-11-12T19:21:12.487

Is it possible to refer directly to s.a as just .a, or similar?

No, .a in this context is a designator, and it cannot be referred to in the brace-or-equal-initializer used to initialize a given data member by means of its matching designator.

struct Bar { int a; };

struct Foo {
    int a;
    Bar b;
};

int main() {
    Foo f = {
     //     ^ start of braced-init-list
        .a
     // ^^ designator (for data member of Foo)
            = 1,
     //     ^^^ initializer-clause
        .b{.a = 2}
     //   ^^^^^^^^ braced-init-list
     //   referring to a designator of a 
     //   data member of Bar
    };
}

Details

Designated initializers, a new C++20 feature introduced by P0329R4, are part of the grammar for braced-init-lists:

braced-init-list:
  { initializer-list ,opt }
  { designated-initializer-list ,opt }
  { }

where the grammar of designated-initializer-list is:

designated-initializer-list:
  designated-initializer-clause
  designated-initializer-list , designated-initializer-clause

and, where the grammar for the individual designated-initializer-clause:s is:

designated-initializer-clause:
  designator brace-or-equal-initializer

and where finally, a designator is:

designator:
  . identifier

Now, a brace-or-equal-initializer does not allow, in any way, to refer to another designator of the same object. Note here the difference here between designators and other identifiers, e.g. referring to data members of the struct as part of the initializer-clause or nested braced-init-list:s of the brace-or-equal-initializer.

Aggregate / designated initialization of c++ struct: Refer directly to another field

1 Answers1

Details