Return with aggregate initialization without type name

user_3234803 0
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In C++ 20, for a type like this:

struct Person {
  std::string name;
  int age;
};

We can return it from a function using aggregate initialization, with or without writing out the type name:

Person getJohn()
{
  // With type name
  return Person {
    .name = "John",
    .age = 42,
  };

  // Without type name
  return {
    .name = "John",
    .age = 42,
  };
}

What's the difference between specifying and not specifying the type name explicitly? One thing I noticed is that without a type name, if I mess up the field names/order, the error message given by my compiler and IDE is less clear than if I have a type name. But other than that, are the two forms semantically different in any way?

c++20 c++
1 Answers

They should always be equivalent. A return statement performs copy-initialization ([stmt.return]/2). A copy-initialization from a designated-initializer-list performs aggregate initialization ([dcl.init.list]/3.1). If the operand is simply the designated-initializer-list itself, we are done. The return object has been aggregate-initialized. If the operand is Person designated-initializer-list, then guaranteed copy elision applies ([dcl.init.general]/16.6.1) which again means that the return object is simply aggregate-initialized (there is no additional temporary object).

Omitting the type can only make a difference in non-aggregate-initialization contexts: if the user-defined constructor selected for the initialization has been declared explicit, the program is ill-formed. But aggregate initialization doesn't use a constructor.

user_4812670
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