自动推导应由操作员或功能返回的类型

Question

模板转换运算符的非常特殊行为的目的是什么？请考虑以下代码：

#include <iostream>
#include <utility>

#include <cstdlib>

#define PP { std::cout << __PRETTY_FUNCTION__ << std::endl; }

struct A
{
    template< typename T >
    operator T () { PP; return {}; }
    template< template< typename ... > class C, typename ...types >
    operator C< types... > () { PP; return {}; }
    template< typename T >
    T operator * () { PP; return {}; }
};

static void f(int) PP

template< typename T >
struct S {};

template< typename T = void >
void g(S< T >) PP

int main()
{
    struct B {};

    // T A::operator B() [T = B]
    B{} = A{};

    // C<types...> A::operator S() [C = S, types = <void>]
    S< void >{} = A{};

    // T A::operator int() [T = int]
    // void f(int)
    f(A{});

    // void g(S<T>) [T = void]
    g({});

    // expected the same output prepended by "T A::operator S() [T = S<void>]"
    // but conversion operator does not involved after first step when matching failed
    // and default value of template parameter should be chosen to next try
    // error: no matching function for call to 'g'
    // note: candidate template ignored: could not match 'S<type-parameter-0-0>' against 'A'
    //g(A{});

    // error: indirection requires pointer operand ('A' invalid)
    //B{} = *A{};

    // error: no matching member function for call to 'operator*'
    // note: candidate template ignored: couldn't infer template argument 'T'
    //B{} = A{}.operator * ();

    // error: indirection requires pointer operand ('A' invalid)
    //(*A{}).~A();
    return EXIT_SUCCESS;
}

就我所见，它能够自动推导出接收者所需的返回类型（赋值表达式的左侧或实际函数参数类型）是唯一的。我无法为一元operator *，operator ->，operator ()和其他人重现此类行为。

附加：

我在隐式转换中找到了trick of how to know预期类型。