This is an unofficial snapshot of the ISO/IEC JTC1 SC22 WG21 Core Issues List revision 120c. See http://www.open-std.org/jtc1/sc22/wg21/ for the official list.
2026-08-20
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1395. Partial ordering of variadic templates reconsidered
Section: 13.10.3.6 [
] Status: C++17 Submitter: John Spicer Date: 2011-09-21 [Moved to DR at the November, 2016 meeting.]
The resolution of
(found in document N3281) made the following example ambiguous and thus ill-formed:
template<class T> void print(ostream &os, const T &t) { os << t; } template <class T, class... Args> void print(ostream &os, const T &t, const Args&... rest) { os << t << ", "; print(os, rest...); } int main() { print(cout, 42); print(cout, 42, 1.23); } This pattern seems fairly intuitive; is it reason to reconsider or modify the outcome of
?
(See also
.)
Notes from the October, 2012 meeting:
CWG agreed that the example should be accepted, handling this case as a late tiebreaker, preferring an omitted parameter over a parameter pack.
Additional note (March, 2013):
For another example:
template<typename ...T> int f(T*...) { return 1; } template<typename T> int f(const T&) { return 2; } int main() { if (f((int*)0) != 1) { return 1; } return 0; } This worked as expected prior to the resolution of
.
Proposed resolution (June, 2016):
Change 13.10.3.5 [
] paragraph 8 as follows:
If A was transformed from a function parameter pack and P is not a parameter pack, type deduction fails. Otherwise, usingUsing the resulting types P and A, the deduction is then done as described in 13.10.3.6 [
]. If P is a function parameter pack, the type A of each remaining parameter type of the argument template is compared with the type P of the declarator-id of the function parameter pack. Each comparison deduces template arguments for subsequent positions in the template parameter packs expanded by the function parameter pack. Similarly, if A was transformed from a function parameter pack, it is compared with each remaining parameter type of the parameter template. If deduction succeeds for a given type, the type from the argument template is considered to be at least as specialized as the type from the parameter template. [Example:...
Add the following as a new paragraph following 13.10.3.5 [
] paragraph 10:
Function template F is at least as specialized as function template G if, for each pair of types used to determine the ordering, the type from F is at least as specialized as the type from G. F is more specialized than G if F is at least as specialized as G and G is not at least as specialized as F.
If, after considering the above, function template F is at least as specialized as function template G and vice-versa, and if G has a trailing paramter pack for which F does not have a corresponding parameter, and if F does not have a trailing parameter pack, then F is more specialized than G.
This resolution also resolves
.