constexpr

I am beginning with c++11, constexpr and template metaprogramming seems a nice way to save scarce ram on tiny microcontroler. Is there a way to write a template to flatten a list of constexpr array, what I need is a way to do : constexpr std::array<int, 3> a1 = {1,2,3}; constexpr std::array<int, 2> a2 = {4,5}; constexpr auto a3 = make_flattened_array (a1,a2); I use gcc 4.8.4 (arm-none-eabi), and can compile with std=c++11 or c++1y option if it is needed. Marco A. Notice - I understood your question as follows: you want to join those two arrays and flatten the result into a single, new array

At the very end of Scott Schurr's talk "Introducing constexpr " at CppCon , he asks "Is there a way to poison a function"? He then explains that this can be done (albeit in a non-standard way) by: Putting a throw in a constexpr function Declaring an unresolved extern const char* Referencing the unresolved extern in the throw I sense that I'm a bit out of my depth here, but I'm curious: What does it mean to "poison a function"? What is the significance/usefulness of the technique he outlines? In general it refers to making a function unusable, e.g. if you want to ban the use of dynamic

In C++11 I am using a constexpr function as a default value for a template parameter - it looks like this: template <int value> struct bar { static constexpr int get() { return value; } }; template <typename A, int value = A::get()> struct foo { }; int main() { typedef foo<bar<0>> type; return 0; } G++ 4.5 and 4.7 compiles this, but Clang++ 3.1 does not. The error message from clang is: clang_test.cpp:10:35: error: non-type template argument is not a constant expression template <typename A, int value = A::get()> ^~~~~~~~ clang_test.cpp:17:19: note: while checking a default template argument

Why is this constexpr static member function, identified by the //! Nah comment, not seen as constexpr when called? struct Item_id { enum Enum { size, position, attributes, window_rect, max_window_size, _ }; static constexpr int n_items_ = _; // OK constexpr auto member_n_items() const -> int { return _; } // OK static constexpr auto static_n_items() -> int { return _; } // OK static constexpr int so_far = n_items_; // OK #ifndef OUT_OF_CLASS static constexpr int bah = static_n_items(); //! Nah. #endif }; constexpr auto n_ids() -> int { return Item_id().member_n_items(); } // OK auto main() ->

I have been looking at the new constexpr feature of C++ and I do not fully understand the need for it. For example, the following code: constexpr int MaxSize() { ... return ...; } void foo() { int vec[MaxSize()]; } can be replaced by: int MaxSize() { ... return ...; } static const int s_maxSize = MaxSize(); foo() { int vec[s_maxSize]; } Update The second example is actually not standard ISO C++ (thanks to several users for pointing this out) but certain compilers (e.g. gcc) support it. So it is not const that makes the program valid, but the fact that gcc supports this non-standard feature.

I am trying to initialize a constexpr reference with no success. I tried #include <iostream> constexpr int& f(int& x) // can define functions returning constexpr references { return x; } int main() { constexpr int x{20}; constexpr const int& z = x; // error here } but I'm getting a compile time error error: constexpr variable 'z' must be initialized by a constant expression Dropping the const results in error: binding of reference to type 'int' to a value of type 'const int' drops qualifiers even though I had the feeling that constexpr automatically implies const for variable declarations. So