Static arrays at compile time in C++

Question

I was wondering if its possible to make the following answer more generic, in the sense that the type of the array be templated instead of just unsigned:

I've enclosed the whole thing in a struct like so:

template<typename ArrayType>
struct Array
{
template<ArrayType... args> struct ArrayHolder {
    static const ArrayType data[sizeof...(args)];
};

template<ArrayType... args> 
const ArrayType ArrayHolder<args...>::data[sizeof...(args)] = { args... };

template<size_t N, template<size_t> class F, ArrayType... args> 
struct generate_array_impl {
    typedef typename generate_array_impl<N-1, F, F<N>::value, args...>::result result;
};

template<template<size_t> class F, ArrayType... args> 
struct generate_array_impl<0, F, args...> {
    typedef ArrayHolder<F<0>::value, args...> result;
};

template<size_t N, template<size_t> class F> 
struct generate_array {
    typedef typename generate_array_impl<N-1, F>::result result;
};
};

but I get the following errors:

c++-4.6 -std=c++0x -o test test.cpp
test.cpp:49:17: error: specializing member ‘Array<ArrayType>::ArrayHolder<args>::data’ requires ‘template<>’ syntax

When showing compiler errors and code, it is usually quite helpfull to point in the code the exact line that the compiler is complaining about (i.e. which is line 49?) — David Rodríguez - dribeas
– David Rodríguez - dribeas, Commented Aug 25, 2011 at 11:20

rodrigo · Accepted Answer · 2011-08-25 10:32:51Z

It helps if you indent the struct. The problem is that you are defining the data static member variable inside the Array struct. But it should be at namespace scope:

template<typename ArrayType>
struct Array
{
    template<ArrayType... args> struct ArrayHolder {
        static const ArrayType data[sizeof...(args)];
    };

    template<size_t N, template<size_t> class F, ArrayType... args> 
        struct generate_array_impl {
            typedef typename generate_array_impl<N-1, F, F<N>::value, args...>::result result;
        };

    template<template<size_t> class F, ArrayType... args> 
        struct generate_array_impl<0, F, args...> {
            typedef ArrayHolder<F<0>::value, args...> result;
        };

    template<size_t N, template<size_t> class F> 
        struct generate_array {
            typedef typename generate_array_impl<N-1, F>::result result;
        };
};

template<typename ArrayType> template<ArrayType... args> 
        const ArrayType Array<ArrayType>::ArrayHolder<args...>::data[sizeof...(args)] = { args... };

Arafangion · Accepted Answer · 2011-08-25 09:45:54Z

-2

Erm... Why don't you just use std::vector? Or if you want a non-resiable array, then use the equivalent type in boost?

answered Aug 25, 2011 at 9:45

Arafangion

12k1 gold badge45 silver badges74 bronze badges

Comments

jj1 · Accepted Answer · 2011-08-25 10:38:01Z

-2

I don't think the expression like 'template ' is a valid usage of C++11 variadic templates.

It can be of two forms as I understand:

template <typename... args> : variable number of generic type parameters
template <int... args> : variable number of integer(non-type) parameters

Refer to this wikipedia article for variadic templates.

answered Aug 25, 2011 at 10:38

jj1

1551 silver badge7 bronze badges

1 Comment

Sebastian Mach Over a year ago

Those are template template arguments and they are valid.

Collectives™ on Stack Overflow

Static arrays at compile time in C++

3 Answers 3

Comments

Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related