Since it is possible that a function declared as constexpr can be called during run-time, under which criteria does the compiler decide whether to compute it at compile-time or during runtime?
template<typename base_t, typename expo_t>
constexpr base_t POW(base_t base, expo_t expo)
{
return (expo != 0 )? base * POW(base, expo -1) : 1;
}
int main(int argc, char** argv)
{
int i = 0;
std::cin >> i;
std::cout << POW(i, 2) << std::endl;
return 0;
}
In this case, i is unknown at compile-time, which is probably the reason why the compiler treats POW() as a regular function which is called at runtime. This dynamic however, as convenient as it may appear to be, has some impractical implications. For instance, could there be a case where I would like the compiler to compute a constexpr function during compile-time, where the compiler decides to treat it as a normal function instead, when it would have worked during compile-time as well? Are there any known common pitfalls?
A constexpr function that is eligible to be evaluated at compile-time will only be evaluated at compile-time if the return value is used where a constant expression is required. Otherwise, compile-time evaluation is not guaranteed.
If you pass a non-constexpr argument to the PlusOne function, the compiler can't evaluate it at compile-time, and it will simply be a normal run-time function. Hint: "The constexpr specifier declares that it is possible to evaluate the value of the function or variable at compile time."
A constexpr function is one whose return value is computable at compile time when consuming code requires it. Consuming code requires the return value at compile time to initialize a constexpr variable, or to provide a non-type template argument.
Yes ([dcl. constexpr], §7.1. 5/2 in the C++11 standard): "constexpr functions and constexpr constructors are implicitly inline (7.1.
constexpr
functions will be evaluated at compile time when all its arguments are constant expressions and the result is used in a constant expression as well. A constant expression could be a literal (like 42
), a non-type template argument (like N
in template<class T, size_t N> class array;
), an enum
element declaration (like Blue
in enum Color { Red, Blue, Green };
, another variable declared constexpr, and so on.
They might be evaluated when all its arguments are constant expressions and the result is not used in a constant expression, but that is up to the implementation.
The function has to be evaluated at compile-time when a constant expression is needed.
The simplest method to guarantee this is to use a constexpr
value, or std::integral_constant
:
constexpr auto result = POW(i, 2); // this should not compile since i is not a constant expression
std::cout << result << std::endl;
or:
std::cout << std::integral_constant<int, POW(i, 2)>::value << std::endl;
or
#define POW_C(base, power) (std::integral_constant<decltype(POW((base), (power)), POW((base), (power))>::value)
std::cout << POW_C(63, 2) << std::endl;
or
template<int base, int power>
struct POW_C {
static constexpr int value = POW(base, power);
};
std::cout << POW_C<2, 63>::value << std::endl;
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