CTAD不适用于默认模板参数？

Question

当我有一个接受向量的类对象时，比较下面的情况。未推导的参数T可以用默认的模板参数很好地替换：

#include <vector>

template <typename T = int>
struct container
{
    container(std::vector<T> vec) {}
};

int main()
{
    container C = std::vector{1,2,3,4,5};
}

对于我的班级来说，情况并非如此，因为情况要复杂一些( struct+container+:+public+obj { ++++container(obj<0>+init)+{} }; int+main() { ++++container<5>+some_container+=+obj{1,2,5,2,obj{1,2,33},2,2}; }'),l:'5',n:'0',o:'C+++source+#1',t:'0')),k:50,l:'4',n:'0',o:'',s:0,t:'0'),(g:!((g:!((h:compiler,i:(compiler:gsnapshot,filters:(b:'0',binary:'1',commentOnly:'0',demangle:'0',directives:'0',execute:'0',intel:'0',libraryCode:'0',trim:'1'),flagsViewOpen:'1',fontScale:14,fontUsePx:'0',j:1,lang:c++,libs:!(),options:'--std=c++17+-O2',selection:(endColumn:1,endLineNumber:1,positionColumn:1,positionLineNumber:1,selectionStartColumn:1,selectionStartLineNumber:1,startColumn:1,startLineNumber:1),source:1,tree:'1'),l:'5',n:'0',o:'x86-64+gcc+(trunk)+(C++,+Editor+#1,+Compiler+#1)',t:'0')),k:50,l:'4',m:43.35736354273945,n:'0',o:'',s:0,t:'0'),(g:!((h:output,i:(compilerName:'x86-64+clang+(trunk)',editorid:1,fontScale:14,fontUsePx:'0',j:1,wrap:'1'),l:'5',n:'0',o:'Output+of+x86-64+gcc+(trunk)+(Compiler+#1)',t:'0')),header:(),l:'4',m:56.64263645726055,n:'0',o:'',s:0,t:'0')),k:50,l:'3',n:'0',o:'',t:'0')),l:'2',n:'0',o:'',t:'0')),version:4))">CompilerExplorer )

#include <cstdio>
#include <initializer_list>
#include <variant>

template <size_t> struct obj;

template<size_t Vs>
using val = std::variant<std::monostate, int, struct obj<Vs>>;

template <size_t Vs = 0>
struct obj
{
    obj() = default;
    obj(std::initializer_list<val<Vs>> init) {
        printf("initializer of object called, Vs = %d\n", Vs);
    }
};


template <size_t Vs = 0>
struct container : public obj<Vs>
{
    container(obj<0> init) {}
};


int main()
{
    container<5> some_container = obj{1,2,5,2,obj{1,2,33},2,2};
}

如果出现以下错误，这将失败：

<source>: In function 'int main()':
<source>:29:57: error: class template argument deduction failed:
   29 |     container<5> some_container = obj{1,2,5,2,obj{1,2,33},2,2};
      |                                                         ^
<source>:29:57: error: no matching function for call to 'obj(int, int, int)'
<source>:14:5: note: candidate: 'template<long unsigned int Vs> obj(std::initializer_list<std::variant<std::monostate, int, obj<Vs> > >)-> obj<<anonymous> >'
   14 |     obj(std::initializer_list<val<Vs>> init) {
      |     ^~~

但是，当我在容器实例化中补充模板专门化obj<0> (在main中)时，它可以工作。你知道为什么这对我的课不起作用吗?我怎样才能解决它？我不想强迫用户每次指定模板。

Answer 1

这个问题在更简单的情况下已经存在。

auto o = obj{1,2,33};

从而产生此错误：

<source>:29:24: error: class template argument deduction failed:
   29 |     auto o = obj{1,2,33};
      |                        ^
<source>:29:24: error: no matching function for call to 'obj(int, int, int)'
<source>:14:5: note: candidate: 'template<long unsigned int Vs> obj(std::initializer_list<std::variant<std::monostate, int, obj<Vs> > >)-> obj<<anonymous> >'
   14 |     obj(std::initializer_list<val<Vs>> init) {
      |     ^~~
<source>:14:5: note:   template argument deduction/substitution failed:
<source>:29:24: note:   mismatched types 'std::initializer_list<std::variant<std::monostate, int, obj<Vs> > >' and 'int'
   29 |     auto o = obj{1,2,33};

因此，编译器无法推断出这三个int应该是一个初始化程序列表。如果在它们周围添加额外的大括号，编译器就会意识到这实际上应该是一个列表参数，而不是三个单独的参数，它可以工作：

auto o = obj{{1,2,33}};

这也涉及到更复杂的情况：

container some_container = obj{{1,2,5,2,obj{{1,2,33}},2,2}};