Which of these new expressions with char arrays are well-formed?

Question

For the following program:

int main() 
{
    new char[4] {"text"};  // #1
    new char[5] {"text"};  // #2
    new char[] {"text"         


        

Answer 1

OK, this is pretty simple to trace. The presence of {} means that list initialization is being performed, so we get to visit our favorite part of the spec: [dcl.init.list]/3.

The object being initialized in case 1 is a char[4]. The braced-init-list is not a designated initializer, so 3.1 is ignored. char[4] is not a class, so 3.2 is ignored. That brings us to 3.3:

Otherwise, if T is a character array and the initializer list has a single element that is an appropriately-typed string-literal ([dcl.init.string]), initialization is performed as described in that subclause.

Well, char[4] is definitely a character array, and the initializer list definitely contains a single element, and that element does in fact match the type of the character array. So off to [dcl.init.string] we go.

That tells us (after a fashion):

Successive characters of the value of the string-literal initialize the elements of the array.

But the next paragraph warns:

There shall not be more initializers than there are array elements.

Well, that makes #1 ill-formed.

So, we redo the process for char[5]. And that doesn't trigger, since 5 is sufficiently large.

Lastly, we come to char[]. Which is no different from using a number, as far as initialization is concerned. char[] is an array of characters, so it follows the above rules. C++17 would choke on using char[] in a new expression, but C++20 is fine with it.

If the type-id or new-type-id denotes an array type of unknown bound ([dcl.array]), the new-initializer shall not be omitted; the allocated object is an array with n elements, where n is determined from the number of initial elements supplied in the new-initializer ([dcl.init.aggr], [dcl.init.string]).

Which means that #2 and #3 are supposed to be legal. So GCC is wrong to make them ill-formed. And it makes #1 ill-formed for the wrong reason.