Kdjykuj

Is it possible to create a double which holds the value of 1*10^x where x is based on a integer template parameter. So something like:

template < int exp >

struct DoubleValue

{

    static constexpr double value = ????;

}



double d = DoubleValue<20>::value; // = 1e20

double d = DoubleValue<-20>::value; // = 1e-20

As it can be created with litterals, it seems that something like this should be possible.

I would like the value to be evaluated at compile time (so std::pow will not work as far as I know).
Also, if possible, I would like to be able to avoid actual iterative computations ((maybe unfounded) fear for precision problems). I would also like to be able to use larger values as exponent, like for example 200, which makes it impossible to store the value in a standerd integer type.

Assuming that your compiler supports C++14 or higher (which should be a valid assumption in the year 2019) this is very simple using a constexpr function:

constexpr double myPow(double x, int exp)

{

    double pow = 1.0;

    for (int i = 0; i < exp; ++i)

        pow *= x;

    for (int i = 0; i > exp; --i)

        pow /= x;

    return pow;

}



template < int exp >

struct DoubleValue

{

    static constexpr double value = myPow(10.0, exp);

};

See here to verify that it works and that even without optimization the value is generated at compile time.

Depending on your use case you might not even need the DoubleValue struct but can directly use myPow().

This one works for me:

// File: 54195854.cpp

// Author: iBug



#include <iostream>

#include <cmath>



template <int e>

struct DoubleValue {

    static double value;

};



template <int e>

double DoubleValue<e>::value = std::pow(10.0, e);



int main(void) {

    std::cout << DoubleValue<10>::value << std::endl;

    std::cout << DoubleValue<-10>::value << std::endl;

}

I don't know if it's correct, but at least the above code compiles with G++ 7.3.0:

g++ -std=c++14 -pedantic 54195854.cpp

And the compiled program a.out generates this output:

1e+10

1e-10

Since you need the value to available in compile time, pretty much the only way to solve it, that came to my mind is recursive templates. However, the fact, that you need for said template to do different things, based on the signedness of the passed value, complicates things. First thing that would come to mind, would be to write such a recursive template:

template <int exp>

struct DoubleValue

    {

    static constexpr double value = (exp < 0

                                     ? DoubleValue<exp+1>::value / 10

                                     : 10 * DoubleValue<exp-1>::value);

    };



// Default case

template <>

struct DoubleValue<0>

    {

    static constexpr double value = 1;

    };

However, such solution wouldn't work, due to the fact, that both branches of the ternary expression, would need to be resolved, and that would, always, lead to the infinite recursion, since one of the branches wouldn't tend to 0. Then, SFINAE came to mind:

// Base case.

template <int exp, class Enable = void>

struct DoubleValue

    {

    };



// Case when exp is positive

template <int exp>

struct DoubleValue<exp, typename std::enable_if<(exp > 0)>::type>

    {

    static constexpr double value = 10 * DoubleValue<exp-1>::value;

    };



// Case when exp is negative

template <int exp>

struct DoubleValue<exp, typename std::enable_if<(exp < 0)>::type>

    {

    static constexpr double value = DoubleValue<exp+1>::value / 10;

    };



// Default case.

template <>

struct DoubleValue<0>

    {

    static constexpr double value = 1;

    };

Live Demo.

If you want it at compile time without std::pow, this should do it:

#include <iostream>



template <int e>

struct DoubleValue {

    static constexpr double value = 10.0 * DoubleValue<e - 1>::value;

};

template <>

struct DoubleValue<0> {

    static constexpr double value = 1.0;

};

int main() {

    std::cout << DoubleValue<20>::value << 'n'; //1e+20

}

C++ Fiddle