In C++11 is sqrt defined as constexpr?

std::sqrt is not defined as constexpr, according to section 26.8 of N3291: the C++11 FDIS (and I doubt they added it to the final standard after that). One could possibly write such a version, but the standard library version is not constexpr.

Just in case anyone is interested in a meta integer square root function, here is one I wrote while a ago:

constexpr std::size_t isqrt_impl
    (std::size_t sq, std::size_t dlt, std::size_t value){
    return sq <= value ?
        isqrt_impl(sq+dlt, dlt+2, value) : (dlt >> 1) - 1;
}

constexpr std::size_t isqrt(std::size_t value){
    return isqrt_impl(1, 3, value);
}

Here is a fast and efficient constexpr implementation for double floating point numbers. You may adapt it to float too, if needed:

#include <limits>   

namespace Detail
{
    double constexpr sqrtNewtonRaphson(double x, double curr, double prev)
    {
        return curr == prev
            ? curr
            : sqrtNewtonRaphson(x, 0.5 * (curr + x / curr), curr);
    }
}

/*
* Constexpr version of the square root
* Return value:
*   - For a finite and non-negative value of "x", returns an approximation for the square root of "x"
*   - Otherwise, returns NaN
*/
double constexpr sqrt(double x)
{
    return x >= 0 && x < std::numeric_limits<double>::infinity()
        ? Detail::sqrtNewtonRaphson(x, x, 0)
        : std::numeric_limits<double>::quiet_NaN();
}