可以在编译时定义一个静态数组，如下：
One can define a static array at compile time as follows:
const std::size_t size = 5; unsigned int list[size] = { 1, 2, 3, 4, 5 };问题1 - 可以使用各种元编程技术在编译时以编程方式分配这些值？
Question 1 - Is it possible by using various kinds of metaprogramming techniques to assign these values "programmatically" at compile time?
问题2 - 假设数组中的所有值都是相同的，是否可以以编程方式在编译时选择性分配值？
Question 2 - Assuming all the values in the array are to be the same barr a few, is it possible to selectively assign values at compile time in a programmatic manner?
例如：
const std::size_t size = 7; unsigned int list[size] = { 0, 0, 2, 3, 0, 0, 0 };
UPDATE: Georg Fritzsche's solution is amazing, needs a little work to get it compiling on msvc and intel compilers, but nonetheless a very interesting approach to the problem.
推荐答案你可以得到的最接近的是使用C ++ 0x功能来初始化本地或成员数组的模板一个可变参数模板参数列表。 这当然是受最大模板实例化深度的限制，实际上会在你的情况下产生显着的差异。
The closest you can get is using C++0x features to initialize local or member arrays of templates from a variadic template argument list. This is of course limited by the maximum template instantiation depth and wether that actually makes a notable difference in your case would have to be measured.
示例：
template<unsigned... args> struct ArrayHolder { static const unsigned data[sizeof...(args)]; }; template<unsigned... args> const unsigned ArrayHolder<args...>::data[sizeof...(args)] = { args... }; template<size_t N, template<size_t> class F, unsigned... 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, unsigned... 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; };1..5 ：
template<size_t index> struct MetaFunc { enum { value = index + 1 }; }; void test() { const size_t count = 5; typedef generate_array<count, MetaFunc>::result A; for (size_t i=0; i<count; ++i) std::cout << A::data[i] << "\n"; }更多推荐
以编程方式在C ++中以编译方式以编程方式创建静态数组
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