Conversion operator template specialization - c++

Question

Here's a largely academic exercise in understanding conversion operators, templates and template specializations. The conversion operator template in the following code works for int, float, and double, but fails when used with std::string... sort of. I've created a specialization of the conversion to std::string, which works when used with initialization std::string s = a;, but fails when used with a cast static_cast<std::string>(a).
#include <iostream>
#include <string>
#include <sstream>
class MyClass {
int y;
public:
MyClass(int v) : y(v) {}
template <typename T>
operator T() { return y; };
};
template<>
MyClass::operator std::string() {
std::stringstream ss;
ss << y << " bottles of beer.";
return ss.str();
}
int main () {
MyClass a(99);
int i = a;
float f = a;
double d = a;
std::string s = a;
std::cerr << static_cast<int>(a) << std::endl;
std::cerr << static_cast<float>(a) << std::endl;
std::cerr << static_cast<double>(a) << std::endl;
std::cerr << static_cast<std::string>(a) << std::endl; // Compiler error
}
The above code generates a compiler error in g++ and icc, both complaining that no user-defined conversion is suitable for converting a MyClass instance to a std::string on the static_cast (C-style casts behave the same).
If I replace the above code with explicit, non-template versions of the conversion operator, everything is happy:
class MyClass {
int y;
public:
MyClass(int v) : y(v) {}
operator double() {return y;}
operator float() {return y;}
operator int() {return y;}
operator std::string() {
std::stringstream ss;
ss << y << " bottles of beer.";
return ss.str();
}
};
What is wrong with my template specialization for std::string? Why does it work for initialization but not casting?
Update:
After some template wizardry by #luc-danton (meta-programming tricks I'd never seen before), I have the following code working in g++ 4.4.5 after enabling experimental C++0x extensions. Aside from the horror of what is being done here, requiring experimental compiler options is reason enough alone to not do this. Regardless, this is hopefully as educational for others as it was for me:
class MyClass {
int y;
public:
MyClass(int v) : y(v) {}
operator std::string() { return "nobody"; }
template <
typename T
, typename Decayed = typename std::decay<T>::type
, typename NotUsed = typename std::enable_if<
!std::is_same<const char*, Decayed>::value &&
!std::is_same<std::allocator<char>, Decayed>::value &&
!std::is_same<std::initializer_list<char>, Decayed>::value
>::type
>
operator T() { return y; }
};
This apparently forces the compiler to choose the conversion operator std::string() for std::string, which gets past whatever ambiguity the compiler was encountering.

Answer 1

You can reproduce the problem by just using
std::string t(a);
Combined with the actual error from GCC (error: call of overloaded 'basic_string(MyClass&)' is ambiguous) we have strong clues as to what may be happening: there is one preferred conversion sequence in the case of copy initialization (std::string s = a;), and in the case of direct initialization (std::string t(a); and static_cast) there are at least two sequences where one of them can't be preferred over the other.
Looking at all the std::basic_string explicit constructors taking one argument (the only ones that would be considered during direct initialization but not copy initialization), we find explicit basic_string(const Allocator& a = Allocator()); which is in fact the only explicit constructor.
Unfortunately I can't do much beyond that diagnostic: I can't think of a trick to discover is operator std::allocator<char> is instantiated or not (I tried SFINAE and operator std::allocator<char>() = delete;, to no success), and I know too little about function template specializations, overload resolution and library requirements to know if the behaviour of GCC is conforming or not.
Since you say the exercise is academic, I will spare you the usual diatribe how non-explicit conversion operators are not a good idea. I think your code is a good enough example as to why anyway :)
I got SFINAE to work. If the operator is declared as:
template <
typename T
, typename Decayed = typename std::decay<T>::type
, typename = typename std::enable_if<
!std::is_same<
const char*
, Decayed
>::value
&& !std::is_same<
std::allocator<char>
, Decayed
>::value
&& !std::is_same<
std::initializer_list<char>
, Decayed
>::value
>::type
>
operator T();
Then there is no ambiguity and the code will compile, the specialization for std::string will be picked and the resulting program will behave as desired. I still don't have an explanation as to why copy initialization is fine.

Answer 2

static_cast here is equivalent of doing std::string(a).
Note that std::string s = std::string(a); doesn't compile either. My guess is, there are plenty of overloads for the constructor, and the template version can convert a to many suitable types.
On the other hand, with a fixed list of conversions, only one of those matches exactly a type that the string's constructor accepts.
To test this, add a conversion to const char* - the non-templated version should start failing at the same place.
(Now the question is why std::string s = a; works. Subtle differences between that and std::string s = std::string(a); are only known to gods.)