C ++中的隐式转换

Question

Given the following code, why doesn't the compiler resolve the implicit conversion when constructing Bar ? That is, construct Foo just like a was constructed which is (should) then be used to construct Bar ?

#include <string>

class ImplicitlyConvertToChar
{
public:
  ImplicitlyConvertToChar(const char* a_char)
    : m_string(a_char)
  { }

  ImplicitlyConvertToChar(const char* a_char, size_t a_end)
    : m_string(a_char)
  {
  }

  template <typename T_String>
  ImplicitlyConvertToChar(T_String const& a_string)
    : m_string(a_string.begin())
  {
  }

  operator char const * () const
  { return m_string; }

  const char* m_string;
};

class Foo
{
public:

  Foo(const ImplicitlyConvertToChar& a_charLike)
    : m_string(a_charLike)
  { }

  const char* m_string;
};

class Bar
{
public:
  Bar(const Foo& a_foo)
    : m_foo(a_foo)
  { }

  Foo m_foo;
};

int main()
{
  Foo a("this works");
  Bar b("Why doesn't this?");
}

Answer 1

You are not allowed more than one user defined implicit conversion. The Foo example involves one, the Bar example involves two.

Answer 2

The compiler is only allowed to make a single implicit user-defined conversion.

See http://en.cppreference.com/w/cpp/language/implicit_cast

A user-defined conversion consists of: 
    zero or one non-explicit single-argument constructor or non-explicit 
    conversion function calls

Constructing Bar that way would require two.