P0846R0
John Spicer <jhs@edg.com>
Audience: CWG
2017-11-08

P0846R0: ADL and Function Templates that are not Visible

Introduction

In Toronto (7/2017) core reviewed Robert Haberlach's D0389R1 "template keyword in unqualified-ids". Core agreed with that the problem presented but disagreed with the intended solution.

Instead of requiring the user to use the template keyword, a revision to the lookup rules was proposed so that a name for which a normal lookup produces either no result or finds one or more functions and that is followed by a a "<" would treated as if a function template name had been found and would cause ADL to be performed.

This proposal was brought to the Evolution group in Toronto and it received strong support (12 | 14 | 0 | 0 | 0).

It was observed that this change could change code where you have an overloaded < operator that accepts a function as the left-hand operand. This case was considered as pathological case not likely enough to be of concern.

Wording changes

Change 6.4.1 [basic.lookup.unquql] paragraph 3:
The lookup for an unqualified name used as the postfix-expression of a function call is described in 6.4.2. [ Note: For purposes of determining (during parsing) whether an expression is a postfix-expression for a function call, the usual name lookup rules apply. In some cases a name followed by < is treated as a template-name even though name lookup did not find a template-name (see 17.2). For example,

int h;
void g();
namespace N {
	struct A {};
	template <class T> int f(T);
	template <class T> int g(T);
	template <class T> int h(T);
}

int x = f<N::A>(N::A());  // OK: lookup of f finds nothing,
                          // f treated as template name
int y = g<N::A>(N::A());  // OK: lookup of g finds a function,
                          // g treated as template name
int z = h<N::A>(N::A());  // error: "h<" does not begin a template-id

The rules in 6.4.2 have no effect on the syntactic interpretation of an expression. For example,
Change 17.2 [temp.names] paragraph 2 and 3:
For a template-name to be explicitly qualified by the template arguments, the name must be known considered to refer to a template. [Note: Whether a name actually refers to a template cannot be known in some cases until after argument dependent lookup is done [6.4.2]. -- end note] A name is considered to refer to a template if name lookup finds a template-name or a overload set that contains a function template. A name is also considered to refer to a template if it is an unqualified-id followed by a < and name lookup finds either one or more functions or finds nothing.

After name lookup (6.4) finds that a name is When a name is considered to be a template-name or that an operator-function-id or a literal- operator-id refers to a set of overloaded functions any member of which is a function template, if this and it is followed by a <, the < is always taken as the delimiter of a template-argument-list and never as the less-than operator.

Change 17.3 [temp.arg] paragraph 7:
When the template name lookup for the name in a template-id is an overloaded function template finds an overload set, both non-template functions in the overload set and function templates in the overload set ...
Add a new section to Annex C.6:

C.6.X [temp.name]

Change: A unqualified-id that is followed by a < and for which name lookup finds nothing or finds a function will be treated as a template-name in order to potentially cause argument dependent lookup to be performed.

Rationale: It was problematic to call a function template with an explicit template argument list via argument dependent lookup because of the need to have a template with the same name visible via normal lookup.

Effect on original feature: Previously valid code that uses a function name as the left operand of a < operator would become ill-formed.


struct A {};
bool operator <(void (*fp)(), A);
void f(){}
int main() {
  A a;
  f < a;  // ill-formed; previously well-formed
  (f) < a;  // still well formed
}