C++ template specialization with <int&> not picking up an int

Question

Hi, I have the following code:

template <typename T> LuaCall& operator>>(T) { BOOST_STATIC_ASSERT(sizeof(T) == 0); }
template <> LuaCall& operator>><int&>(int& val) { mResults.push_back(std::make_pair(LUA_RESULT_INTEGER, (void *)&val)); return *this; }
template <> LuaCall& operator>><float&>(float& val) { mResults.push_back(std::make_pair(LUA_RESULT_FLOAT, (void *)&val)); return *this; }
template <> LuaCall& operator>><double&>(double& val) { mResults.push_back(std::make_pair(LUA_RESULT_DOUBLE, (void *)&val)); return *this; }
template <> LuaCall& operator>><bool&>(bool& val) { mResults.push_back(std::make_pair(LUA_RESULT_BOOLEAN, (void *)&val)); return *this; }
template <> LuaCall& operator>><std::string&>(std::string& val) { mResults.push_back(std::make_pair(LUA_RESULT_STRING, (void *)&val)); return *this; }
template <> LuaCall& operator>><LuaNilStruct>(LuaNilStruct) { mResults.push_back(std::make_pair(LUA_RESULT_NIL, (void *)NULL)); return *this; }

And then:

int abc;
LuaCall(l, "test") % "test" % 5 % LuaNil % 2.333 >> abc;

I want it to work kinda like cin >> does, ie it needs to write to abc the return value of the lua function. So I need its address.. but it defaults on the default template. What am I doing wrong? There is surely a way to do this since cin does exactly that.

Thanks!

---

Note to whoever changed the %'s to >>: I changed it back since it's the way it is :D The code calls the Lua function test("test", 5, nil, 2.333) and saves its return value to abc. %'s are for the parameters of the functions, >>'s are for the return value(s).

template <typename T>
LuaCall& operator%(T val) {
 mLua->Push(val);
 ++mArguments;
 return *this;
}

Answer 1

For one, operator>> is a binary operator. That means it has to take two arguments. The first argument has to be whatever is the type of

LuaCall(l, "test") % "test" % 5 % LuaNil % 2.333

Since that operator returns LuaCall&, I suppose that's what the result of that call (and thus left argument) is.

From that mResult I suppose that you attempt to do that operator as a class member. But I don't think you can specialize member templates.

Finally, I don't think you should specialize for reference types. This

template <> LuaCall& operator>><int>(int val);

should do

Answer 2

You'v written operator>> as a unary operator, but it's a binary operator. LeftHandSide >> RightHandSide.

The form std::cout <"Hello" << " world"; therefore is (operator<<(operator<<(std::cout, "Hello"), " world); - the first << returns std::cout for use as the left-hand side of the second <<.

The real problem is that in your code, lookup happens as follows:

1. Candiate functions are determined (only one candidate, the template)
2. T is deduced.
3. Determine which specialization (if any) to instantiate.

In step 2, T==int. In step 3, there are no specializations for T==int so the base template is chosen and instantiated. You might want to use an overload instead of a specialization. The overload would be a better match in step 1 so you don't even get to the point of template argument deduction

Answer 3

You cannot use constant values ("test" , 5 or 2.333) where references are expected. Change the template and parameter type of your operator>> to (int, float etc.) when you want this behaviour.