Why if constexpr fails to bypass constexpr evaluation? - c++

Question

I'm building a static loop for type dispatching using macros. Here is
what I achieved so far.
#define LOOP(n, f) \
static_assert(n <= 8 && "static loop size should <= 8"); \
do { \
if constexpr (n >= 8) \
f(std::integral_constant<size_t, n - 8>()); \
if constexpr (n >= 7) \
f(std::integral_constant<size_t, n - 7>()); \
if constexpr (n >= 6) \
f(std::integral_constant<size_t, n - 6>()); \
if constexpr (n >= 5) \
f(std::integral_constant<size_t, n - 5>()); \
if constexpr (n >= 4) \
f(std::integral_constant<size_t, n - 4>()); \
if constexpr (n >= 3) \
f(std::integral_constant<size_t, n - 3>()); \
if constexpr (n >= 2) \
f(std::integral_constant<size_t, n - 2>()); \
if constexpr (n >= 1) \
f(std::integral_constant<size_t, n - 1>()); \
} while (0);
template <typename T> constexpr size_t tupleSize(T&) { return tuple_size_v<T>; }
int main() {
auto t = std::make_tuple(1, "string", 0.2, 3, 1, 1, 1);
LOOP(tupleSize(t), [&](auto i) { cout << std::get<i>(t) << endl; });
return 0;
}
And the godbolt link https://godbolt.org/z/GcMZI3
The question is, why do the first four branches fail the compilation?

Answer 1

Do not use a macro, use a function template instead. if constexpr works by discarding the non-taken branch depending on the current instantiation of a template.
template <std::size_t n, typename F>
void loop(F&& f)
{
static_assert(n <= 8 && "static loop size should <= 8");
if constexpr (n >= 8)
f(std::integral_constant<size_t, n - 8>());
if constexpr (n >= 7)
f(std::integral_constant<size_t, n - 7>());
if constexpr (n >= 6)
f(std::integral_constant<size_t, n - 6>());
if constexpr (n >= 5)
f(std::integral_constant<size_t, n - 5>());
if constexpr (n >= 4)
f(std::integral_constant<size_t, n - 4>());
if constexpr (n >= 3)
f(std::integral_constant<size_t, n - 3>());
if constexpr (n >= 2)
f(std::integral_constant<size_t, n - 2>());
if constexpr (n >= 1)
f(std::integral_constant<size_t, n - 1>());
}
Usage:
int main() {
constexpr auto t = std::make_tuple(1, "string", 0.2, 3);
loop<tupleSize(t)>([&](auto i) { cout << std::get<i>(t) << endl; });
return 0;
}
live example on godbolt.org
From cppreference:
If a constexpr if statement appears inside a templated entity, and if condition is not value-dependent after instantiation, the discarded statement is not instantiated when the enclosing template is instantiated.
Outside a template, a discarded statement is fully checked. if constexpr is not a substitute for the #if preprocessing directive