I have the following very simple template. As I learned, ^ is not the exponential operator. Now I'm looking for a way to compute this power. There are many examples with a recursive template on the internet. This is not too difficult.
But I wonder: Is there actually no "built-in" method in C++ to compute this on compile time?
template <int DIM>
class BinIdx : Idx
{
static const int SIZE = 3 ^ DIM; // whoops, this is NOT an exponential operator!
}
As mentioned you can use << if the exponent is a power of two.
Otherwise, if the exponents are non-negative integers you can write a constexpr function like this one.
template<typename T, typename U>
auto constexpr pow(T base, U exponent) {
static_assert(std::is_integral<U>(), "exponent must be integral");
return exponent == 0 ? 1 : base * pow(base, exponent - 1);
}
This will obviously break for large exponents as well as negative ones, though.
I am not fully aware of how well compilers optimize function calls in constant expressions. Here's a manual optimization for cases where the exponents are powers of two. This will also reduce the amount of recursion done.
template<typename T>
bool constexpr is_power_of_two(T x) {
return (x != 0) && ((x & (x - 1)) == 0);
}
template<typename T, typename U>
auto constexpr pow(T base, U exponent) {
static_assert(std::is_integral<U>(), "exponent must be integral");
if (is_power_of_two(exponent)) {
return base << exponent;
}
return exponent == 0 ? 1 : base * pow(base, exponent - 1);
}
More efficient algorithms are also available. However, I am bad at computer science so I don't know how to implement them.
You can use template metaprogramming. Let me show the code.
template <int A, int B>
struct get_power
{
static const int value = A * get_power<A, B - 1>::value;
};
template <int A>
struct get_power<A, 0>
{
static const int value = 1;
};
Usage:
std::cout << get_power<3, 3>::value << std::endl;
As an addition to elyse's answer, here is a version with recursion depth of log(n):
template<typename T>
constexpr T sqr(T a) {
return a * a;
}
template<typename T>
constexpr T power(T a, std::size_t n) {
return n == 0 ? 1 : sqr(power(a, n / 2)) * (n % 2 == 0 ? 1 : a);
}
No, there is no general built in way to calculate the power of values. There is the pow function from the standard library and you can use the << shift operator for the special case 2^x.
This would work in your case (*):
static const int SIZE = (1 << DIM);
* = You updated your question from 2^x to 3^x after I wrote my answer.
For another special case x^y where x and y are static, you can just write a long multiplication:
const result int = x*x*x*x*x;
A named operator library:
namespace named_operator {
template<class D>struct make_operator{
constexpr make_operator(){}
};
template<class T, char, class O> struct half_apply { T&& lhs; };
template<class Lhs, class Op>
constexpr
half_apply<Lhs, '*', Op>
operator*( Lhs&& lhs, make_operator<Op> ) {
return {std::forward<Lhs>(lhs)};
}
template<class Lhs, class Op, class Rhs>
constexpr auto
times( Lhs&& lhs, Op, Rhs&& rhs, ... ) // ... keeps this the worst option
-> decltype( invoke( std::declval<Lhs>(), Op{}, std::declval<Rhs>() ) )
{
// pure ADL call, usually based off the type Op:
return invoke( std::forward<Lhs>(lhs), Op{}, std::forward<Rhs>(rhs) );
}
template<class Lhs, class Op, class Rhs>
constexpr auto
operator*( half_apply<Lhs, '*', Op>&& lhs, Rhs&& rhs )
-> decltype(
times( std::declval<Lhs>(), Op{}, std::declval<Rhs>() )
)
{
return times( std::forward<Lhs>(lhs.lhs), Op{}, std::forward<Rhs>(rhs) );
}
}
It only supports operator*, but extending it should be obvious. Picking names for times equivalents is a bit of an issue.
@Anton's solution, augmented with a named operator:
namespace power {
template<typename T>
constexpr T sqr(T a) {
return a * a;
}
template<typename T>
constexpr T power(T a, std::size_t n) {
return n == 0 ? 1 : sqr(power(a, n / 2)) * (n % 2 == 0 ? 1 : a);
}
namespace details {
struct pow_tag {};
constexpr named_operator::make_operator<pow_tag> pow;
template<class Scalar>
constexpr Scalar times( Scalar lhs, pow_tag, std::size_t rhs ) {
return power( std::forward<Scalar>(lhs), rhs );
}
}
using details::pow;
}
and now this works:
using power::pow;
int array[ 2 *pow* 10 ] = {0};
