Empty variardic packs of enums -- do they make two functions different? - c++

There is a technique I sometimes use when overriding template functions that goes like this:
#include <utility>
template<int> struct unique_enum { enum class type {}; };
template<int index> using UniqueEnum = typename unique_enum<index>::type;
template<bool b, int index=1>
using EnableFuncIf = typename std::enable_if< b, UniqueEnum<index> >::type;
template<bool b, int index=1>
using DisableFuncIf = EnableFuncIf<!b, -index>;
// boring traits class:
template<typename T>
struct is_int : std::false_type {};
template<>
struct is_int<int> : std::true_type {};
#include <iostream>
// use empty variardic packs to give these two SFINAE functions different signatures:
template<typename C, EnableFuncIf< is_int<C>::value >...>
void do_stuff() {
std::cout << "int!\n";
}
template<typename C, DisableFuncIf< is_int<C>::value >...>
void do_stuff() {
std::cout << "not int!\n";
}
int main() {
do_stuff<int>();
do_stuff<double>();
}
This distinguishes do_stuff from do_stuff, because one takes 0 or more UniqueEnum<1>s, and the other takes 0 or more UniqueEnum<-1>s. gcc 4.8 considers these different empty packs to be distinct.
However, in the latest version of clang I tried, this fails: it treats the function with 0 UniqueEnum<1>s as being the same as the function with 0 UniqueEnum<-1>s.
There are easy workarounds that work in clang, but I'm wondering if my above technique is legal -- do two function templates, which differ only by empty variardic parameter packs, actually different?

I think GCC is right, and your technique is correct. Basically, since the type argument for C is specified explicitly, the question is whether:
a. substitution of C everywhere else in the function template signature happens first, and then type deduction is performed (which should result in a substitution failure); or
b. type deduction is performed first, and then substitution is performed (which would not result in a substitution failure, because the corresponding argument pack would be empty, and so there would be no substitution to perform).
It seems GCC assumes (1), while Clang assumes (2). Paragraph 14.8.2/2 of the C++11 Standard specifies:
When an explicit template argument list is specified, the template arguments must be compatible with the
template parameter list and must result in a valid function type as described below; otherwise type deduction
fails. Specifically, the following steps are performed when evaluating an explicitly specified template
argument list with respect to a given function template:
— The specified template arguments must match the template parameters in kind (i.e., type, non-type,
template). There must not be more arguments than there are parameters unless at least one parameter
is a template parameter pack, and there shall be an argument for each non-pack parameter. Otherwise,
type deduction fails.
— Non-type arguments must match the types of the corresponding non-type template parameters, or must
be convertible to the types of the corresponding non-type parameters as specified in 14.3.2, otherwise
type deduction fails.
— The specified template argument values are substituted for the corresponding template parameters as
specified below.
The following paragraph then says:
After this substitution is performed, the function parameter type adjustments described in 8.3.5 are performed. [...]
Moreover, paragraph 14.8.2/5 specifies:
The resulting substituted and adjusted function type is used as the type of the function template for template
argument deduction. [...]
Finally, paragraph 14.8.2/6 goes as follows:
At certain points in the template argument deduction process it is necessary to take a function type that
makes use of template parameters and replace those template parameters with the corresponding template
arguments. This is done at the beginning of template argument deduction when any explicitly specified template
arguments are substituted into the function type, and again at the end of template argument deduction
when any template arguments that were deduced or obtained from default arguments are substituted.
This all seems to imply that first substitution is performed, then template argument deduction. Hence, a substitution failure should occur in either case and one of the two templates should be discarded from the overload set.
Unfortunately, there does not seem to be a clear specification as to what the behavior should be when templates arguments are deduced rather than being explicitly specified.

Related

Exact rules for matching variadic template template parameters in partial template specialization

While creating this answer for another question I came around the following issue. Consider this program (godbolt):
#include <variant>
#include <iostream>
template <typename T>
struct TypeChecker {
void operator()() {
std::cout << "I am other type\n";
}
};
template<typename... Ts, template<typename...> typename V>
requires std::same_as<V<Ts...>, std::variant<Ts...>>
struct TypeChecker<V<Ts...>>
{
void operator()()
{
std::cout << "I am std::variant\n";
}
};
int main()
{
TypeChecker<std::variant<int, float>>{}();
TypeChecker<int>{}();
}
The output (which is also expected) is the following (with clang 14.0.0 as well as with gcc 12.1):
I am std::variant
I am other type
If however the three dots in the parameter list of the template template are removed, like this (whole program live on godbolt):
template<typename... Ts, template<typename> typename V>
,then the output is different for clang and gcc. The clang 14.0.0 compiled program outputs
I am other type
I am other type
whereas the gcc 12.1 compiled program outputs
I am std::variant
I am other type
It seems that using the non-variadic template template exhibits different matching rules in clang and gcc. So my question is, which behavior is correct if it is even well defined, and why?
Since defect report resolution P0522R0 was adopted, exact matching of template parameter lists for template template parameter match is no longer needed, and correct output according to the standard is:
I am std::variant
I am other type
In the current draft (which also contains changes related to C++20 concepts) relevant standard excerpts are temp.arg.template#3-4 (bold emphasis mine):
A template-argument matches a template template-parameter P when P is at least as specialized as the template-argument A. In this comparison, if P is unconstrained, the constraints on A are not considered. If P contains a template parameter pack, then A also matches P if each of A's template parameters matches the corresponding template parameter in the template-head of P. Two template parameters match if they are of the same kind (type, non-type, template), for non-type template-parameters, their types are equivalent ([temp.over.link]), and for template template-parameters, each of their corresponding template-parameters matches, recursively. When P's template-head contains a template parameter pack ([temp.variadic]), the template parameter pack will match zero or more template parameters or template parameter packs in the template-head of A with the same type and form as the template parameter pack in P (ignoring whether those template parameters are template parameter packs).
A template template-parameter P is at least as specialized as a template template-argument A if, given the following rewrite to two function templates, the function template corresponding to P is at least as specialized as the function template corresponding to A according to the partial ordering rules for function templates. Given an invented class template X with the template-head of A (including default arguments and requires-clause, if any):
(4.1) Each of the two function templates has the same template parameters and requires-clause (if any), respectively, as P or A.
(4.2) Each function template has a single function parameter whose type is a specialization of X with template arguments corresponding to the template parameters from the respective function template where, for each template parameter PP in the template-head of the function template, a corresponding template argument AA is formed. If PP declares a template parameter pack, then AA is the pack expansion PP... ([temp.variadic]); otherwise, AA is the id-expression PP.
If the rewrite produces an invalid type, then P is not at least as specialized as A.
So, as we see, exact (up to special rules for parameter pack) parameter matching is now considered only in the case parameter list of template template parameter contains a pack (like your original example), otherwise only the new at least as specialized as relation is used to test matching, which defines for both template template parameter and argument respective function templates and tests whether parameter-induced function is at least as specialized as argument-induced function according to partial ordering rules for template functions.
In particular, matching A=std::variant to P=template<typename> typename V we get these corresponding function templates:
template<typename...> class X;
template<typename T> void f(X<T>); // for P
template<typename... Ts> void f(X<Ts...>); // for A
So, to prove A matches P, we need to prove f(X<T>) is at least as specialized as f(X<Ts...>). temp.func.order#2-4 says:
Partial ordering selects which of two function templates is more specialized than the other by transforming each template in turn (see next paragraph) and performing template argument deduction using the function type. The deduction process determines whether one of the templates is more specialized than the other. If so, the more specialized template is the one chosen by the partial ordering process. If both deductions succeed, the partial ordering selects the more constrained template (if one exists) as determined below.
To produce the transformed template, for each type, non-type, or template template parameter (including template parameter packs thereof) synthesize a unique type, value, or class template respectively and substitute it for each occurrence of that parameter in the function type of the template. ...
Using the transformed function template's function type, perform type deduction against the other template as described in [temp.deduct.partial].
Transformed template of f(X<T>) is f(X<U1>) and of f(X<Ts...>) is f(X<U2FromPack>), where U1 and U2FromPack are two synthesized unique types. Now, temp.deduct.partial#2-4,8,10 says:
Two sets of types are used to determine the partial ordering. For each of the templates involved there is the original function type and the transformed function type.
[Note 1: The creation of the transformed type is described in [temp.func.order]. — end note]
The deduction process uses the transformed type as the argument template and the original type of the other template as the parameter template. This process is done twice for each type involved in the partial ordering comparison: once using the transformed template-1 as the argument template and template-2 as the parameter template and again using the transformed template-2 as the argument template and template-1 as the parameter template.
The types used to determine the ordering depend on the context in which the partial ordering is done:
(3.1) In the context of a function call, the types used are those function parameter types for which the function call has arguments.130
(3.2) In the context of a call to a conversion function, the return types of the conversion function templates are used.
(3.3) In other contexts the function template's function type is used.
Each type nominated above from the parameter template and the corresponding type from the argument template are used as the types of P and A.
Using the resulting types P and A, the deduction is then done as described in [temp.deduct.type]. ... 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.
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.
Now, in our case, per 3.3, the function type itself is the only one considered among types to determine ordering. So, per 2, 8 and 10, to know whether f(X<T>) is at least as specialized as f(X<Ts...>) we need to see whether void(X<T>) is at least as specialized as void(X<Ts...>), or, equivalently, whether deduction from type for P=void(X<Ts...>) from A=void(X<U1>) succeeds. It does according to temp.deduct.type#9-10:
If P has a form that contains <T> or <i>, then each argument Pi
of the respective template argument list of P is compared with the corresponding argument Ai of the corresponding template argument list of A. If the template argument list of P contains a pack expansion that is not the last template argument, the entire template argument list is a non-deduced context. If Pi is a pack expansion, then the pattern of Pi is compared with each remaining argument in the template argument list of A. Each comparison deduces template arguments for subsequent positions in the template parameter packs expanded by Pi. ...
Similarly, if P has a form that contains (T), then each parameter type Pi of the respective parameter-type-list ([dcl.fct]) of P is compared with the corresponding parameter type Ai of the corresponding parameter-type-list of A. ...
Here, per 10, comparison of functions types results in a single comparison of X<Ts...> and X<U1>, which, according to 9, succeeds.
Thus, deduction is succesful, so f(X<T>) is indeed at least as specialized as f(X<Ts...>), so std::variant matches template<typename> typename V. Intuitively, we gave 'more general' template template argument which should work nicely for intended usage of a more specific template template parameter.
In practice, different compilers enable P0522R0 changes under different circumstances, and cppreference template parameters page (section Template template arguments) contains links and information on GCC, Cland and MSVC. In particular, GCC in C++17+ mode enables it by default (and for previous standards with compiler flag fnew-ttp-matching), but Clang doesn't in any mode unless -frelaxed-template-template-args flag is provided, thus you got the difference in output for them. With the flag, Clang also produces correct behaviour (godbolt).

deducing function template specialization when taking address

I'm trying to understand the rules for function template argument deduction in the case where all arguments are defaulted. Under 13.10.1 (Explicit template argument specification), the standard (C++20) says:
— when the address of a function is taken, when a function initializes a reference to function, or when a pointer to member function is formed, ... If all of the template arguments can be deduced, they may all be omitted; in this case, the empty template argument list <> itself may also be omitted.
However, in the fourth line of the code snippets below, compilers (gcc, MSVC) seem to insist on having the empty angle brackets. Is this non-conformance or have I missed something ? (In this example, the issue doesn't matter, but the question arose in a context where it does matter).
template <typename T = int> void f(T) {}
void (*p)(int) = f; //ok
auto a = f<>; //ok
auto b = f; //error, can't deduce
void g() {}
auto c = g; //OK
The compilers are correct: auto a = f<>; is well-formed and auto b = f; is ill-formed. This is due to the rules about auto, combined with the rules about the address of the function template name f.
As described in [dcl.type.auto.deduct]/4, the auto type for the variable definitions is found in a way like template argument deduction done for a hypothetical function template using a template type parameter in its function parameter:
template <typename U> void auto_deducer(U);
auto a = f<>; // auto becomes the U deduced for auto_deducer(f<>)
auto b = f; // auto becomes the U deduced for auto_deducer(f)
In [temp.deduct.type] paragraphs 4 and 5:
In certain contexts, however, the value [of a template argument] does not participate in type deduction, but instead uses the values of template arguments that were either deduced elsewhere or explicitly specified. If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.
The non-deduced contexts are:
...
A function parameter for which the associated argument is an overload set, and one or more of the following apply:
...
the overload set supplied as an argument contains one or more function templates.
...
So for both a and b, the template argument deduction used to determine the type of auto fails. Since it's not true that "all of the template arguments can be deduced", we're not allowed to omit the empty <> syntax.
But we're not entirely done yet, because the same paragraph [temp.arg.explicit]/4 you quoted has some relevant additional text:
Trailing template arguments that can be deduced or obtained from default template-arguments may be omitted from the list of explicit template-arguments. A trailing template parameter pack.... If all of the template arguments can be deduced, they may all be omitted; in this case, the empty template argument list <> itself may also be omitted. In contexts where deduction is done and fails, or in contexts where deduction is not done, if a template argument list is specified and it, along with any default template arguments, identifies a single function template specialization, then the template-id is an lvalue for the function template specialization.
For the definition of both a and b, the f<> or f expression is in a non-deduced context, so "deduction is not done". (The template argument deduction for the hypothetical auto_deducer(f<>) then fails, but that's for the overall auto_deducer call, after the part actually involving f<> determined that the type deduction step should not be done with that argument at all.) In auto a = f<>;, "a template argument list is specified" and using the default template argument "identifies a single function template specialization" f<int>, so the paragraph's final sentence applies and the template-id names f<int> after all. In auto b = f; no template argument list is specified (and f is not a template-id), so the sentence can't apply.
An actual call like the statement f(); is fine because template argument deduction happens, uses the default template argument, and succeeds without the complication of auto. Converting expression f to a specific target pointer-to-function type is fine, because that will deduce the template parameter T from the target type, and the default template argument doesn't get involved.
Another confirmation this behavior is intended is found in a non-normative note in [over.over]/3. The [over.over] section applies when lookup for a name (like f or f<>) gives an overload set comprising one or more functions and/or function templates and the name expression is not followed by a function call argument list:
For each function template designated by the name, template argument deduction is done, and if the argument deduction succeeds, the resulting template argument list is used to generate a single function template specialization, which is added to the set of selected functions considered. [ Note: As described in [temp.arg.explicit], if deduction fails and the function template name is followed by an explicit template argument list, the template-id is then examined to see whether it identifies a single function template specialization. If it does, the template-id is considered to be an lvalue for that function template specialization. The target type is not used in that determination. - end note ]
Trailing template arguments that can be deduced or obtained from default template-arguments
may be omitted from the list of explicit template-arguments. [...] If all of the template
arguments can be deduced, they may all be omitted; in this case, the empty template argument list <>
itself may also be omitted.
Note the last sentence, as opposed to the first sentence, does not say "deduced or obtained from default template-arguments". Deducing and obtaining from default template-arguments are two different things. This may or may not be a wording defect in the standard.

Why does the compiler try to deduce when an argument-less function template is provided?

template <typename... T>
struct X {
static_assert(sizeof...(T) != 0);
};
template <typename... T>
void f(const X<T...> &) {}
template <typename T>
void inner() {}
int main() {
f(inner);
}
The static assert fires in this example.
Why does the compiler try to deducing anything here? And then it apparently even tries to instantiate X (with empty type argument list?..)...
Why the error isn't just 'template used without arguments'?..
If I change inner function to be a struct, the compiler reports:
use of class template 'inner' requires template arguments
which makes sense; if the param is just const X &, there's
declaration type contains unexpanded parameter pack 'T'
which also makes sense, however is less clear than the case with struct, because it reports an issue at the callee, not at the call site.
If the param is const X<T> &, the report is also a bit weird at first sight:
candidate template ignored: couldn't infer template argument 'T'.
These errors are for Clang 14, but GCC also reports similar ones.
Is the instantiation here somehow specified by the standard? If so, how? Also, why does it result in an empty type list?
This is mostly due to [temp.arg.explicit]/4, applicable when source names a function template, and emphasis mine:
Trailing template arguments that can be deduced or obtained from default template-arguments may be omitted from the list of explicit template-arguments. A trailing template parameter pack not otherwise deduced will be deduced as an empty sequence of template arguments. If all of the template arguments can be deduced, they may all be omitted; in this case, the empty template argument list <> itself may also be omitted.
The function template name as an argument means it is not used for template argument deduction of f ([temp.deduct.type]/(5.5.3)). So the pack T is not deduced from the function argument list (inner), and [temp.arg.explicit]/4 applies and deduces T as an empty list of types.
Now to evaluate the expression f(inner) involves converting the expression inner to the parameter type const X<>&. Whether and how this is valid depends on the constructors of class X<>, so the template is instantiated, causing the static_assert error since the template parameter pack does have zero elements.
Because the argument refers to an overload set that contains a function template X, no deduction is attempted from it. The hope is that the deduction will succeed anyway (perhaps from other arguments) and then the template arguments of X can be deduced from the resulting argument type. (Of course, it’s impossible to deduce the template argument for inner, but even that of
template<class T>
T make();
can be deduced in certain contexts, so it’s not in general a vain hope.)
Here, deduction for f does succeed vacuously, with T as an empty pack. (This is much like a default template argument.) Then const X<>& is the parameter type, and so overload resolution is attempted to construct an X<> from inner. That obviously depends on the constructors for X<>, so that type is completed and the compilation fails.

What is the normative rule in the current draft rules that a template parameter pack which does not participate in deduction is deduced to empty

template<typename ...T, typename U>
void fun(U){}
int main(){
fun(0);
}
This snippet code is accepted by both GCC and Clang. The template parameter pack T does not participate in the template argument deduction in the context of function call, as per the following rules:
[temp.deduct.call]
Template argument deduction is done by comparing each function template parameter type (call it P) that contains template-parameters that participate in template argument deduction with the type of the corresponding argument of the call (call it A) as described below.
The pack T is contained by any function template parameter. If there were no other special rules specify, the deduction would fail according to:
[temp.deduct.type#2]
if any template argument remains neither deduced nor explicitly specified, template argument deduction fails.
However, such a case is ruled by the following rule in the current standard, that is:
[temp.arg.explicit#4]
A trailing template parameter pack ([temp.variadic]) not otherwise deduced will be deduced as an empty sequence of template arguments.
So, the above case can be considered to deduce successfully which leaves the pack T with an empty set of template arguments.
However, the special rule in temp.arg.explicit#4 has been changed to a note in the current draft
[temp.arg.explicit#note-1]
[Note 1: A trailing template parameter pack ([temp.variadic]) not otherwise deduced will be deduced as an empty sequence of template arguments. — end note]
So, I wonder Is there any alternative normative rule in the current draft states that the pack T not otherwise deduced will be deduced to an empty set of template arguments?
The previously normative section of [temp.arg.explicit]/4
[...] A trailing template parameter pack ([temp.variadic]) not otherwise deduced will be deduced as an empty sequence of template arguments. [...]
was made into a non-normative note as part of P1787R6.
As you've pointed out, as per [temp.deduct.type]/2 [emphasis mine]:
Type deduction is done independently for each P/A pair [...], if any template argument remains neither deduced nor explicitly specified, template argument deduction fails.
[temp.arg.general] describes that a template parameter that is a template parameter pack may correspond to zero template arguments:
[...] When the parameter declared by the template is a template parameter pack, it will correspond to zero or more template-arguments.
and [temp.variadic]/1 explicitly mention that a template parameter pack may accept zero arguments:
A template parameter pack is a template parameter that accepts zero or more template arguments.
followed by a non-normative example of an empty argument list for an entity templated over a parameter pack template parameter:
template<class ... Types> struct Tuple { };
Tuple<> t0; // Types contains no arguments
Now, returning to [temp.arg.explicit]/4:
If all of the template arguments can be deduced, they may all be omitted; in this case, the empty template argument list <> itself may also be omitted.
Meaning the Tuple example above is can likewise omit the empty argument list
Tuple t0; // Also OK: Types contains no arguments
but where the key is that, as per [temp.arg.general] above, that a template parameter list may correspond to zero arguments, in which case there are no template arguments needed to be deduced.
If you look at your own example:
template<typename ...T, typename U>
void fun(U){}
int main(){
fun(0); // #1
}
you could likewise invoke #1 as:
fun<>(0); // argument list for parameter pack is empty
// -> no argument (beyond that for `U`) to deduce
highlighting that the deducible argument corresponding to the template parameter U can be omitted from the explicit template-arguments, whereas the remaining template-arguments are none; namely, the argument list for the template parameter that is a template parameter pack is empty, and there are thus no remaining template arguments that needs to be deduced.
Thus
A trailing template parameter pack ([temp.variadic]) not otherwise deduced will be deduced as an empty sequence of template arguments.
is non-normative/redundant, explaining why it was wrapped into a [Note - [...] - end note].

How does the template argument deduction perform for function template parameter when it is a class template with default argument

template<typename T, typename U = T>
struct Test{};
template<typename T>
void func(Test<T>){ //#1
}
int main(){
func(Test<int>{}); //#2
}
Consider the above code, At the point of invocation of function template func, the type of argument is Test<int,int>, When call the function template, template argument deduction will perform.
The rule of template argument deduction for function call is :
temp.deduct#call-1
Template argument deduction is done by comparing each function template parameter type (call it P) that contains template-parameters that participate in template argument deduction with the type of the corresponding argument of the call (call it A) as described below.
I'm pretty sure the type of A is Test<int,int>, however I'm not sure what the type of P here is. Is it Test<T> or Test<T,T>, According to the rule, It seems to the type of P here is Test<T>, then deduction process is performed to determine the value of T that participate in template argument deduction. Then according to these rules described as the following:
temp.deduct#call-4
In general, the deduction process attempts to find template argument values that will make the deduced A identical to A (after the type A is transformed as described above).
temp.deduct#5
When all template arguments have been deduced or obtained from default template arguments, all uses of template parameters in the template parameter list of the template and the function type are replaced with the corresponding deduced or default argument values.
Because the class template Test has a default argument, hence the deduced T is substituted into default argument. That means the deduced A is Test<int,int> and it is identical to Argument type Test<int,int>.
However, It's just my understanding. I'm not sure what type the P here is. If change the type of function argument to Test<int,double>, the outcome will report:
candidate template ignored: deduced conflicting types for parameter 'T' ('int' vs. 'double')
The outcome looks like as if the P is Test<T,T> and the fist value of T is conflicting with the second value of T.
So, My question is:
Whether the P here is Test<T> or Test<T,T>? and why?
not a language lawyer answer
There is no type Test<T> is actually a "shorthand" for Test<T, T>.
Just like with default function arguments if you have int foo(int a, int b = 24) the type of the function is int (int, int) and any call like foo(11) is actually foo(11, 24).
P must be a type not a template. test <T> is a template-id, but it is not explicitly said in the standard that the template-id test <T> is equivalent to test<T,T>. The only thing that is said is:
A template-id is valid if
[...]
there is an argument for each non-deducible non-pack parameter that does not have a default template-argument, [...]
After that, holes in the standard are filled by our intuition oriented by the use of the term default.
I think the key point here is that a template designate a family, and a template-id cannot designate a family.
Whether the P here is Test<T> or Test<T,T>? and why?
P is Test<T,T>.
I think we can agree that the rules of [temp.deduct] applies also for class templates; e.g. [temp.class.order], covering partial ordering of class template specializations, is entirely based on the concept of re-writing the class templates to (invented) function templates and applying the rules of function templates to that of the invented function templates corresponding to the original class templates under partial ordering analysis. Combined with the fact that the standard passage for class templates is quite brief in comparison to function templates, I interpret the references below as applying also for class templates.
Now, from [temp.deduct]/1 [emphasis mine]:
When a function template specialization is referenced, all of the template arguments shall have values. The values can be explicitly specified or, in some cases, be deduced from the use or obtained from default template-arguments. [...]
and, from [temp.deduct]/2 [emphasis mine]:
When an explicit template argument list is specified, the template arguments must be compatible with the template parameter list and must result in a valid function type as described below; otherwise type deduction fails. Specifically, the following steps are performed when evaluating an explicitly specified template argument list with respect to a given function template:
(2.1) The specified template arguments must match the template parameters in kind (i.e., type, non-type, template). There must not be more arguments than there are parameters unless [...]
With extra emphasis on "is referenced" and "the specified template arguments"; there is no requirement that we specify all arguments for a given matching function(/class) template, only that those that do specify follow the requirements of [temp.deduct]/2 for explicitly specified template arguments.
This leads us to back to [temp.deduct]/1 for the remaining template arguments of a given candidate function/class template: these can be either deduced (function templates) or obtained from the default template arguments. Thus, the call:
func(Test<int>{});
is, as per the argument above, semantically equivalent to
func(Test<int, int>{});
with the main difference that the template arguments for the former is decided by both an explicitly specified template arguments and a default template argument, whereas for the latter both are decided by explicitly specified template arguments. From this, it is clear that A is Test<int, int>, but we will use a similar argument for P.
From [temp.deduct.type]/3 [emphasis mine]:
A given type P can be composed from a number of other types, templates, and non-type values:
[...]
(3.3) A type that is a specialization of a class template (e.g., A<int>) includes the types, templates, and non-type values referenced by the template argument list of the specialization.
Notice that the description in [temp.deduct.type]/3.3 now returns to the template argument list of the template type P. It doesn't matter that P, for when inspecting this particular candidate function in overload resolution, refers to a class template by partly explicitly specifying the template argument list and partly relying on a default template parameter, where the latter is instantiation-dependent. This step of overload resolution does not imply any kind of instantiation, only inspection of candidates. Thus, the same rules as we just applied to the template argument A above applies to P, in this case, and as Test<int, int> is referenced (via Test<int>), P is Test<int, int>, and we have a perfect match for P and A (for the single parameter-argument pair P and A of this example)
Compiler error messages?
Based in the argument above, one could arguably expect a similar error message for the OP's failing example:
// (Ex1)
template<typename T, typename U = T>
struct Test{};
template<typename T>
void func(Test<T>) {}
int main() {
func(Test<int, double>{});
}
as for the following simple one:
// (Ex2)
struct Foo {};
template<typename T> struct Test {};
template<typename T> void f(T) {}
int main() {
f<Test<int>>(Test<Foo>{});
}
This is not the case, however, as the former yields the following error messages for GCC and Clang, respectively:
// (Ex1)
// GCC
error: no matching function for call to 'func(Test<int, double>)'
note: template argument deduction/substitution failed:
deduced conflicting types for parameter 'T' ('int' and 'double')
// Clang
error: no matching function for call to 'func'
note: candidate template ignored: deduced
conflicting types for parameter 'T' ('int' vs. 'double')
whereas the latter yields the following error messages for GCC and Clang, respectively:
// (Ex2)
// GCC
error: could not convert 'Test<Foo>{}' from 'Test<Foo>' to 'Test<int>'
// Clang
error: no matching function for call to 'f'
note: candidate function template not viable:
no known conversion from 'Test<Foo>' to 'Test<int>' for 1st argument
We can finally note that if we tweak (Ex1) into explicitly specifying the single template argument of f, both GCC and Clang yields similar error messages as for (Ex2), hinting that argument deduction has been entirely removed from the equation.
template<typename T, typename U = T>
struct Test{};
template<typename T>
void func(Test<T>) {}
int main() {
func<int>(Test<int, double>{});
}
The key for this difference may be as specified in [temp.deduct]/6 [emphasis mine]:
At certain points in the template argument deduction process it is necessary to take a function type that makes use of template parameters and replace those template parameters with the corresponding template arguments. This is done at the beginning of template argument deduction when any explicitly specified template arguments are substituted into the function type, and again at the end of template argument deduction when any template arguments that were deduced or obtained from default arguments are substituted.
namely that the template argument deduction process is separated into a clear beginning and end, categorizing:
explicitly specified template arguments as the beginning of the process, and,
deduced or default argument-obtained template arguments as the end of the process,
which would explain the differences in the error messages of the examples above; if all template arguments have been explicitly specified in the beginning of the deduction process, the remainder of the process will not have any remaining template argument to work with w.r.t. deduction or default template arguments.
I tryed to come up with a code that forces only class deduction without function deduction.
Here, there are no function instantiations, but the compiler emits an error anyway:
template<typename T, typename U = T>
struct Test{};
template<typename T>
void func(Test<T, T>){
}
template<typename T>
void func(Test<T>){
}
redefinition of 'template<class T> void func(Test<T, T>)'
GCC: https://godbolt.org/z/7c981E
Clang:
https://godbolt.org/z/G1eKTx
Previous wrong answer:
P refers to template parameter, not to template itself. In declaration Test<typename T, typename U = T> P refers to T, not to Test. So in the instantiation Test<int> T is int, just like A in the call is also int.