我需要创建一个包含范围内所有值的数组。创建后无法设置值，因为数组必须是 constexpr。
template<int FIRST, int LAST>
struct Foo
{
static constexpr int array[LAST - FIRST + 1] = ???;
};
例如Foo<3, 7>::array;
应该等同于static constexpr int array[5] = {3, 4, 5, 6, 7}; .
有可能吗？
最佳答案
是的，它可以在 C++11 中完成。不，它不漂亮。您基本上需要重新实现 C++14 的编译时整数序列。
///////////////// Reimplement compile-time integer sequences ///////////////////////
// There's plenty of better implementations around.
// This one is similar to libstdc++'s implementation.
template<class T, T... Ints>
struct integer_seq {
using next = integer_seq<T, Ints..., sizeof...(Ints)>;
static constexpr std::size_t size() { return sizeof...(Ints); }
};
template<class T, int Len>
struct seq_builder{
static_assert(Len > 0, "Length must be nonnegative");
using type = typename seq_builder<T, Len-1>::type::next;
};
template<class T>
struct seq_builder<T, 0>{
using type = integer_seq<T>;
};
template<class T, int length>
using make_int_sequence = typename seq_builder<T, length>::type;
/////////////////// Actual stuff starts here/////////////////////////////////
template<int FIRST, int LAST, class = make_int_sequence<int, LAST+1-FIRST>>
struct Foo;
template<int FIRST, int LAST, int... SEQ>
struct Foo<FIRST, LAST, integer_seq<int, SEQ...>>
{
static constexpr int array[sizeof...(SEQ)] = {(FIRST+SEQ)...};
};
template<int FIRST, int LAST, int... SEQ>
constexpr int Foo<FIRST, LAST, integer_seq<int, SEQ...>>::array[sizeof...(SEQ)];
Demo .
关于c++ - 按范围初始化数组，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25594551/