N4527 14.8.2.4 [temp.deduct.partial]
3 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.
(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 (14.5.6.2) the function template’s function type is used.
4 Each type nominated above from the parameter template and the corresponding type from the argument template are used as the types of
P
andA
.8 If
A
was transformed from a function parameter pack andP
is not a parameter pack, type deduction fails. Otherwise, using the resulting typesP
andA
, the deduction is then done as described in 14.8.2.5. IfP
is a function parameter pack, the typeA
of each remaining parameter type of the argument template is compared with the typeP
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. 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:
template<class... Args> void f(Args... args); // #1
template<class T1, class... Args> void f(T1 a1, Args... args); // #2
template<class T1, class T2> void f(T1 a1, T2 a2); // #3
f(); // calls #1
f(1, 2, 3); // calls #2
f(1, 2); // calls #3; non-variadic template #3 is more
// specialized than the variadic templates #1 and #2
为什么 f(1, 2, 3);
调用 #2?
我需要更多详细信息,包括:
1 是哪个语境?
2 转化的from是什么?
比如#1 的转换形式是 void (U)
, void (U...)
还是其他形式?(U
表示唯一类型)
14.5.6.2 [temp.func.order]/p3
To produce the transformed template, for each type, non-type, or template template parameter (including template parameter packs (14.5.3) 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.
3 推导中使用的P
和A
有哪些类型?例如
template <class T> void f(T);
int a = 1;
f(a);//P = T, A = int
最佳答案
Why
f(1, 2, 3);
calls #2?
你的问题有很多问题(请每个问题一个问题!),所以我会坚持那个问题。首先,我们执行模板推导。 #3 失败,但 #1 和 #2 成功:
template<class... Args>
void f(Args... args); // #1, with Args = {int, int, int}
template<class T1, class... Args>
void f(T1 a1, Args... args); // #2, with T1 = int, Args = {int, int}
这两个函数的值都采用三个 int
,因此重载决议中的所有正常决胜局都无法解决歧义。所以我们到了最后一个:
Given these definitions, a viable function
F1
is defined to be a better function than another viable functionF2
if for all argumentsi
, ICSi(F1
) is not a worse conversion sequence than ICSi(F2
), and then
— [...]
—F1
andF2
are function template specializations, and the function template forF1
is more specialized than the template forF2
according to the partial ordering rules described in 14.5.6.2.
规则是:
第一步:综合类型[temp.func.order]:
To produce the transformed template, for each type, non-type, or template template parameter (including template parameter packs (14.5.3) 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.
所以我们有:
void f(Pack1... args); // #1
void f(U2 a1, Pack2... args); // #2
第 2 步:按照 [temp.deduct.partial] 中的描述执行扣除。我们所在的上下文是函数调用,因此我们使用函数调用具有参数的类型。
首先,我们尝试从#1 推导出#2。也就是说,我们尝试将 (T1, Args...
) 与 (Pack1...)
进行匹配。第一部分是 P = T1, A = Pack1...
。我们有:
If A was transformed from a function parameter pack and P is not a parameter pack, type deduction fails.
因此从#1 推导#2 失败,因此参数 Args...
至少不像 T1, Args...
那样专门化。
接下来,我们尝试从#2 推导出#1。也就是说,我们尝试将 (Args...)
与 (U2, Pack2...)
进行匹配。成功了,所以 T1, Args...
至少和 Args...
一样专业。
由于#2 至少与#1 一样专业,而#1 至少不如#2 专业,我们可以说#2 更专业:
Function template
F
is at least as specialized as function templateG
if, for each pair of types used to determine the ordering, the type fromF
is at least as specialized as the type fromG
.F
is more specialized thanG
ifF
is at least as specialized asG
andG
is not at least as specialized asF
.
首选更专业的模板,因此我们调用#2。
关于c++ - 当参数是函数参数包时,在部分排序期间推导模板参数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32373979/