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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;
}
Andreas Bonini
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  • Ah - thought you changed your mind halfway through since I didn't spot any suitable operator% in your code. Please post complete code next time. – MSalters Oct 27 '09 at 12:22
  • Indeed, this is the reason why overloading operators to have non-standard meanings is a bad practise. It just leads to misunderstandings when someone else reads your code. – Gorpik Jan 21 '10 at 10:27

3 Answers3

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

MSalters
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1

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.

Frank Bollack
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  • That wasn't supposed to be >>, it's % that does something else, someone edited my question replacing the %'s with >>'s. the only >> is >> abc – Andreas Bonini Oct 27 '09 at 12:26
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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

sbi
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  • This gives an error: Error 1 error C2912: explicit specialization; 'LuaLib::LuaCall &LuaLib::LuaCall::operator >>(int &)' is not a specialization of a function template e:\dev\lualib\inc\luapp.h 161 – Andreas Bonini Oct 27 '09 at 12:24
  • oh, also, if I try operator>> with two arguments it gives an error saying it expects only 1 – Andreas Bonini Oct 27 '09 at 12:28
  • Then the code you've provided is STILL incomplete, because the error message shows us that your operator>> is a member function but in the question it's a free function! – MSalters Oct 27 '09 at 12:32