C++ "forgetting" that variable is constexpr when used as function argument

Question

I have the following code where I am irritated by the fact that compiler is unable to see that variable passed as argument to a function is constexpr so I must use arity 0 function instead of 1 argument function.

I know this is not a compiler bug, but I wonder if there are idioms that enable to workaround this problem.

#include <array>
#include <iostream>

static constexpr std::array<int, 5> arr{11, 22, 33, 44, 55};

template <typename C, typename P, typename Y>
static constexpr void copy_if(const C& rng, P p, Y yi3ld) {
    for (const auto& elem: rng) {
        if (p(elem)){
            yi3ld(elem);
        }
    }
}

// template<std::size_t N>
static constexpr auto get_evens(/* const std::array<int, N>& arr */) {
    constexpr auto is_even = [](const int i) constexpr {return i % 2 == 0;};
    constexpr int cnt = [/* &arr, */&is_even]() constexpr {
        int cnt = 0;
        auto increment = [&cnt] (const auto&){cnt++;};
        copy_if(arr, is_even, increment);
        return cnt;
    }();
    std::array<int, cnt> result{};
    int idx = 0;
    copy_if(arr, is_even, [&result, &idx](const auto& val){ result[idx++] = val;});
    return result;
}

int main() {
    // constexpr std::array<int, 5> arr{11, 22, 33, 44, 55};
    for (const int i:get_evens(/* arr */)) {
        std::cout << i << " " << std::endl;
    }
}

If it is not obvious what I want: I would like to change get_evens signature so that it is template templated on array size N and that it takes 1 argument of type const std::array<int, N>&.

The error message when I change arr to be an function argument isn't helpful:

prog.cc:25:21: note: initializer of 'cnt' is not a constant expression prog.cc:19:19: note: declared here constexpr int cnt = [&arr, &is_even]()constexpr {

Answer 1

#include <array>
#include <iostream>

static constexpr std::array<int, 5> arr{11, 22, 33, 44, 55};

template <typename C, typename P, typename T>
static constexpr void invoke_if(const C& rng, P p, T target) {
    for (const auto& elem: rng) {
        if (p(elem)){
            target(elem);
        }
    }
}

constexpr bool is_even(int i) {
    return i % 2 == 0;
}

template<std::size_t N>
constexpr std::size_t count_evens(const std::array<int, N>& arr)
{
    std::size_t cnt = 0;
    invoke_if(arr, is_even, [&cnt](auto&&){++cnt;});
    return cnt;
}

template<std::size_t cnt, std::size_t N>
static constexpr auto get_evens(const std::array<int, N>& arr) {
    std::array<int, cnt> result{};
    int idx = 0;
    invoke_if(arr, is_even, [&result, &idx](const auto& val){ result[idx++] = val;});
    return result;
}

int main() {
    // constexpr std::array<int, 5> arr{11, 22, 33, 44, 55};
    for (const int i:get_evens<count_evens(arr)>(arr)) {
        std::cout << i << " " << std::endl;
    }
}

this works in g++, but in clang we get a problem because the begin on an array isn't properly constexpr with at least one library. Or maybe g++ violates the standard and clang does not.

Answer 2

template<auto t0, auto...ts>
struct ct_array:
  std::array<decltype(t0) const, 1+sizeof...(ts)>,
  std::integer_sequence<decltype(t0), t0, ts...>
{
  ct_array():std::array<decltype(t0) const, 1+sizeof...(ts)>{{t0, ts...}} {};
};

template<class target, auto X>
struct push;
template<auto X>
struct push<void, X>{using type=ct_array<X>;};
template<auto...elems, auto X>
struct push<ct_array<elems...>, X>{using type=ct_array<elems...,X>;};
template<class target, auto X>
using push_t= typename push<target, X>::type;

template<class target>
struct pop;
template<auto x>
struct pop<ct_array<x>>{using type=void;};
template<auto x0, auto...xs>
struct pop<ct_array<x0, xs...>>{using type=ct_array<xs...>;};
template<class target>
using pop_t=typename pop<target>::type;

template<class lhs, class rhs, class F, class=void>
struct transcribe;
template<class lhs, class rhs, class F>
using transcribe_t = typename transcribe<lhs, rhs, F>::type;

template<auto l0, auto...lhs, class rhs, class F>
struct transcribe<ct_array<l0, lhs...>, rhs, F,
  std::enable_if_t<F{}(l0) && sizeof...(lhs)>
>:
  transcribe<pop_t<ct_array<l0, lhs...>>, push_t<rhs, l0>, F>
{};
template<auto l0, auto...lhs, class rhs, class F>
struct transcribe<ct_array<l0, lhs...>, rhs, F,
  std::enable_if_t<!F{}(l0) && sizeof...(lhs)>
>:
  transcribe<pop_t<ct_array<l0, lhs...>>, rhs, F>
{};
template<auto lhs, class rhs, class F>
struct transcribe<ct_array<lhs>, rhs, F, void>
{
  using type=std::conditional_t< F{}(lhs), push_t<rhs, lhs>, rhs >;
};
template<class lhs, class F>
using filter_t = transcribe_t<lhs, void, F>;

// C++20
//auto is_even = [](auto i)->bool{ return !(i%2); };
struct is_even_t {
  template<class T>
  constexpr bool operator()(T i)const{ return !(i%2); }
};
constexpr is_even_t is_even{};

template<auto...is>
static constexpr auto get_evens(ct_array<is...>) {
  return filter_t< ct_array<is...>, decltype(is_even) >{};
}

Live example.

Test code:

auto arr = ct_array<11, 22, 33, 44, 55>{};
for (const int i : get_evens(arr)) {
    std::cout << i << " " << std::endl;
}

Answer 3

Following the Evg's suggestion, so passing the numbers as template parameters of a std::integer_sequence, but passing the integer sequence as argument of the get_evens() function, and not as template parameter, you can use the numbers directly inside get_evens().

I mean... you can simplify the get_evens() as follows (EDIT: further simplified following a suggestion from Evg (Thanks!))

template <typename T, T ... Ts>
constexpr auto get_evens (std::integer_sequence<T, Ts...> const &)
 {
   std::array<T, (std::size_t(!(Ts & T{1})) + ...)> result{};

   std::size_t idx = 0;

   ((void)(Ts & 1 || (result[idx++] = Ts, true)), ...);

   return result;
 } 

and you can use it this way

int main()
 {
   using arr = std::integer_sequence<int, 11, 22, 33, 44, 55>;

   for ( const int i : get_evens(arr{}) )
      std::cout << i << " " << std::endl;
 }

Answer 4

A function argument is never a constant expression, even if a function is used in constexpr context:

constexpr int foo(int i)
{
    // i is not a constexpr
    return i + 1;
}

constexpr auto i = 1;
constexpr auto j = foo(i);    

To mimic a constexpr argument, use a template parameter:

template<int i>
constexpr int foo()
{
    // i is constexpr
    return i + 1;
}

constexpr auto i = 1;
constexpr auto j = foo<i>();

A possible solution is to use std::integer_sequence to encode integers into a type:

#include <array>
#include <iostream>
#include <type_traits>

template<typename P, typename Y, int... elements>
constexpr void copy_if_impl(P p, Y yi3ld, std::integer_sequence<int, elements...>) {
    ((p(elements) && (yi3ld(elements), true)), ...);
}

template<typename arr_t, typename P, typename Y>
constexpr void copy_if(P p, Y yi3ld) {
    copy_if_impl(p, yi3ld, arr_t{});
}

template<typename arr_t>
constexpr auto get_evens(){
    constexpr auto is_even = [](const int i) constexpr { return i % 2 == 0; };
    constexpr int cnt = [&is_even]() constexpr {
        int cnt = 0;
        auto increment = [&cnt](const auto&) { cnt++; };
        copy_if<arr_t>(is_even, increment);
        return cnt;
    }();

    std::array<int, cnt> result{};
    int idx = 0;
    copy_if<arr_t>(is_even, [&result, &idx](const auto& val) {
        result[idx++] = val; });
    return result;
}

int main()
{
    using arr = std::integer_sequence<int, 11, 22, 33, 44, 55>;
    for (const int i : get_evens<arr>()) {
        std::cout << i << " " << std::endl;
    }
}

Addition suggested by Constantinos Glynos.

From Effective Modern C++ book by Scott Meyers, item 15, p.98:

• constexpr functions can be used in contexts that demand compile-time constants. If the values of the arguments you pass to a constexpr function in such a context are known during compilation, the result will be computed during compilation. If any of the arguments’ values is not known during compilation, your code will be rejected.
• When a constexpr function is called with one or more values that are not known during compilation, it acts like a normal function, computing its result at runtime. This means you don’t need two functions to perform the same operation, one for compile-time constants and one for all other values. The constexpr function does it all.

Answer 5

The other answer has a correct work around but I think the reasoning has nothing to do with parameters but instead to do with the lambda capture here:

constexpr int cnt = [/* &arr, */&is_even]() 

Indeed we can test the various scenarios with this code:

#include <array> 
#include <iostream>

template <size_t N>
constexpr int foo(const std::array<int, N>& arr) {
    return [&arr] () { return arr.size(); }();
}

template <size_t N>
constexpr int bar(const std::array<int, N>& arr) {
    int res{};
    for (auto i : arr) {
        res++;
    }
    return res;
}

template <size_t N>
constexpr int baz(const std::array<int, N>& arr)     {
    constexpr int test = [&arr] () constexpr {
        return bar(arr);
    }();
    return test;
}

int main() {
    constexpr std::array<int, 5> arr{11, 22, 33, 44, 55};
    constexpr std::array<int, foo(arr)> test{};
    constexpr std::array<int, bar(arr)> test2{};
    constexpr std::array<int, baz(arr)> test3{};
}   

Note that the line where test3 is initialized fails to compile. This, however, compiles just fine:

template <size_t N>
constexpr int baz(const std::array<int, N>& arr) {
    return bar(arr);
}

So, what's the problem here? Well lambdas are really just glorified functors, and internally it'll look something like this:

struct constexpr_functor {
    const std::array<int, 5>& arr;
    constexpr constexpr_functor(const std::array<int, 5>& test)
        : arr(test) { }
    constexpr int operator()() const {
        return bar(arr);
    }
};
// ...
constexpr constexpr_functor t{arr};
constexpr std::array<int, t()> test3{};

Notice now that we get an error message showing the real problem:

test.cpp:36:33: note: reference to 'arr' is not a constant expression
test.cpp:33:34: note: declared here
    constexpr std::array<int, 5> arr{11, 22, 33, 44, 55};

The other answer quotes Scotts Meyer's book but misinterprets the quotes. The book actually shows several examples of parameters being used in constexpr situations, but the quotes are simply saying that if you pass a non-constexpr parameter, the function can run at compile-time.