Issue 0035.01: Can one declare an enumeration or struct tag as part of an old-style parameter declaration?

This issue has been automatically converted from the original issue lists and some formatting may not have been preserved.

Authors: Derek M. Jones, WG14
Date: 1992-12-10
Reference document: X3J11/91-039
Submitted against: C90
Status: Closed
Converted from: dr.htm, dr_035.html

void f(a, b)
 int a(enum b {x, y});
 int b;
 {
 }

Now this example is perverse because a prototype declaration is used to declare the parameter of an old-style function declaration. But anyway ...

Is the declaration of the parameter a legal or a constraint error?

Now a(...) is a declarator.

Subclause 6.7.1 says on page 82, lines 7-8:

... each declaration in the declaration list shall have at least one declarator, and those declarators shall declare only identifiers from the identifier list.

The identifier list contains a and b.

The declarator for parameter a declares the identifiers a, b, x, and y.

b is in the identifier list, so that is okay. But x and y are not. Constraint error (methinks so)?

See subclause 6.1.2, page 19 for a definition of an identifier.

Comment from WG14 on 1997-09-23:

Response

There is no constraint violation. The scopes of b, x, and y end at the right-parenthesis at the end of the enum, so there is no violation. It is difficult to call the function f, but there is no constraint violation. The phrase “each declarator declares one identifier” in subclause 6.5.4 refers to a, not to b, x, or y.

As an example, in the conforming definition:

void f(a, b)
 int a(enum b{x, y});
 int b;
 {
 }

the scope of b (the enum tag), x, and y ends at the right-parenthesis at the end of the enum (prototype scope).