Compile time type checking C++

问题

I have created a type list. I then create a class using a template passing the type list. When I call the print function of the class with a some types not specified they are casted. How can I enforce the exact type at compile time? So if I use an unlisted type I get a compiler error. Thanks.

template <class T, class U>
struct Typelist
{
   typedef T Head;
   typedef U Tail;
};


class NullType
{
};

typedef Typelist<int,Typelist<float,Typelist<char*,NullType> > > UsableTypes;

template<class T>
class MyClass
{
    public:
    void print(T::Head _Value) { std::cout << _Value; }
    void print(T::Tail::Head _Value) { std::cout << _Value; }
    void print(T::Tail::Tail::Head _Value) { std::cout << _Value; }
    private:
};


MyClass<UsableTypes> testclass;

void TestMyClass()
{
    int int_val = 100000;
    float flt_val = 0.1f;
    char* char_val = "Hi";
    short short_val = 10;
    std::string str_val = "Hello";

    testclass.print( int_val ); // OK  8-)
    std::cout << endl;
    testclass.print( flt_val ); // OK  8-)
    std::cout << endl;
    testclass.print( char_val ); // OK  8-)
    std::cout << endl;
    testclass.print( short_val); // this compiles OK and works ???  8-(
    std::cout << endl;
    testclass.print( str_val ); // compile error  8-)
    std::cout << endl;
}

@Kerrek SB: Hi I thought it was going to help me with my next step, which was creating the print function depending on the t_list contents, Types and amounts of types. But I'm struggling to separate compile time processing and runtime processing. What I am trying to do is create a print function for each type in the list. So if the list has two types, two print functions will be created and if there are five types then five print functions will be created one for each type. When I do this:

 typedef Typelist<int,Typelist<float,Typelist<char*,NullType> > > UsableTypes;

 MyClass<UsableTypes> newclass

Does this create three instance of MyClass one for each type in the list or does it create one instance and I have to create a print function for each type? I feel I almost have all the blocks in my mind but just can’t fit them together. Any help you can offer would be gratefully received. Thanks.

回答1:

Add a private function template

template<typename T> void print(T);

which doesn't need an implementation. This should catch all types for which no explicit print exists, and since it is private, it will give an error message.

回答2:

You would have to make your print function into a template and then check whether the types match:

template <typename U>
void print(const U & u)
{
  // use std::is_same<typename std::decay<T::Head>::type, typename std::decay<U>::type>::value
}

Here I'm stealing is_same and decay from <type_traits>, but if you don't have C++11, you can either take them from TR1 or from Boost, or just write them yourself, as they're very simple type modifier classes.

The conditional would best go into a static_assert, which is another C++11 feature, but there exist similar constructions for C++98/03 that produce a compile-time error under a certain condition.

回答3:

You could take your arguments by non-const reference, forcing them to be the exact same type. However you can no longer use it with const variables or literals.

来源：https://stackoverflow.com/questions/8106560/compile-time-type-checking-c

标签

c++

templates

casting