Constexpr - function cannot be used in constant expression

Question

I want to compute a lookup table at compile time for a mathematical function in a given range and then retrieve the values ​​from the table at runtime. My code is as follows:

#include <iostream>
#include <cmath>


template<int size>
class LookupTable {
public:
constexpr LookupTable(double xMin, double xMax) : array(), xMin(xMin), xMax(xMax), dx((xMax - xMin) / (size - 1)) {
    for(auto i = 0; i < size; ++i)
        array[i] = exp(xMin + i * dx);
}

constexpr double operator()(double x) const {
    return array[std::min(std::max(static_cast<int>((x - xMin) / dx), 0), size-1)];
}

private:
double array[size];
double xMin;
double xMax;
double dx;
};



int main() {
    const double x = 0.5;
    constexpr LookupTable<10000> table(0.0, 1.0);
    std::cout << "f(x) = " << table(x) << std::endl;  // evaluated at compile time ?
    std::cout << "f(x) = " << LookupTable<10000>(0.0, 1.0)(x) << std::endl;  // evaluated at run time ?
    return 0;
}

The code compiles and runs on gcc 5.1 and higher, but not on Clang 3.8. Clang: constexpr variable 'table' error messages must be initialized with a constant expression, and the non-constexpr 'exp' function cannot be used in a constant expression.

When I remove constexpr in:

constexpr LookupTable<10000> table(0.0, 1.0);

then the code compiles and runs on Clang.

My questions:

• What is the meaning of the "exp" function error message cannot be used in a constant expression "? And is there a workaround?
• , LookupTable < 10000 > (0.0, 1.0); constexpr, ? - , .
• . , ?

Answer 1

gcc, exp() constexpr <cmath>:

17.6.5.6 constexpr [constexpr.functions]

1 , constexpr (7.1.5). - constexpr, , required. , constexpr .

constexpr LookupTable . - static, .

constexpr, , , , <cmath> constexpr.