------- Comment #2 from schaub-johannes at web dot de 2010-03-28 19:33 -------
I've done some analysis for this using the argument-deduction during partial
ordering rules as clarified by
http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#214:
template<typename T, typename U>
void f(U&) { } // #1
template<typename T, typename U>
void f(T const&) { } // #2
4 Each type from the parameter template and the corresponding type from the
argument template are used as the types of P and A.
Round 1, #1 -> #2:
As: (X&) // #1
Ps: (T const&) // #2
Round 2, #2 -> #1:
As: (X const&) // #2
Ps: (T&) // #1
If P is a reference type, P is replaced by the type referred to.
If A is a reference type, A is replaced by the type referred to.
=>
Round 1, #1 -> #2:
As: (X) // #1
Ps: (T const) // #2
Round 2, #2 -> #1:
As: (X const) // #2
Ps: (T) // #1
If both P and A were reference types (before being replaced with the type
referred to above), determine which of the two types (if any) is more
cv-qualified than the other; otherwise the types are considered to be
equally cv-qualified for partial ordering purposes. The result of this
determination will be used below.
=> ************* #2 more cv-qualified than #1
If P is a cv-qualified type, P is replaced by the cv-unqualified
version of P.
If A is a cv-qualified type, A is replaced by the cv-unqualified
version of A.
Round 1, #1 -> #2:
As: (X) // #1
Ps: (T) // #2
Round 2, #2 -> #1:
As: (X) // #2
Ps: (T) // #1
Using the resulting types P and A the deduction is then done as described in
14.9.2.5. 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.
Round 1, T <- X
Round 2, T <- X
#1.param1 at least as specialized as #2.param1 and vice-versa.
If, for a given type, deduction succeeds in both directions (i.e., the types
are identical after the transformations above) and if the type from the
argument template is more cv-qualified than the type from the parameter
template (as described above) that type is considered to be more specialized
than the other.
Deduction succeeded in both rounds ("types" as in P/A pairs), but #2.param1
was more cv-qualified
=> #2.param1 more specialized than #1.param1
If for each type being considered a given template is at least as specialized
for all types and more specialized for some set of types and the other template
is not more specialized for any types or is not at least as specialized for any
types, then the given template is more specialized than the other template.
#2 is at least as specialized for all types and more specialized for #1.param1
and #1 is not more specialized for any types, so #2 is more specialized than
#1.
So, i think GCC should choose the one with the f(T const&) because that
template is more specialized. Notice that the rules also say
In most cases, all template parameters must have values in order for deduction
to succeed, but for partial ordering purposes a template parameter may remain
without a value provided it is not used in the types being used for partial
ordering.
--
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=43559
