In C++11 is sqrt defined as constexpr?

Question

In C++11 is std::sqrt defined as constexpr, i.e. can it legally be used from other constexpr functions or in compile-time contexts like array sizes or template arguments? g++ seems to allow it (using -std=c++0x), but I'm not sure I can take that as authoritative given that c++0x/c++11 support is still incomplete. The fact that I can't seem to find anything on the Internet makes me unsure.

It seems like this should be something one could easily find out using Google, but I've tried (for 40 minutes now...) and couldn't find anything. I could find several proposals for adding constexpr to various parts of the standard library (like for example this one), but nothing about sqrt or other math functions.

Answer 1

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.

Answer 2

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);
}

Answer 3

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();
}

Answer 4

Below is a constexpr square root implementation that uses binary search. It works correctly up to 2^64 with gcc and clang, other simpler versions often fail for numbers > 2^32 because compilers limit the recursion depth to e.g. 200.

// C++11 compile time square root using binary search

#define MID ((lo + hi + 1) / 2)

constexpr uint64_t sqrt_helper(uint64_t x, uint64_t lo, uint64_t hi)
{
  return lo == hi ? lo : ((x / MID < MID)
      ? sqrt_helper(x, lo, MID - 1) : sqrt_helper(x, MID, hi));
}

constexpr uint64_t ct_sqrt(uint64_t x)
{
  return sqrt_helper(x, 0, x / 2 + 1);
}

Below is a nicer version (for integer constants) which requires C++14, it is similar to the one presented in Baptiste Wicht's blog post. C++14 constexpr functions are allowed to use local variables and if statements.

// C++14 compile time square root using binary search

template <typename T>
constexpr T sqrt_helper(T x, T lo, T hi)
{
  if (lo == hi)
    return lo;

  const T mid = (lo + hi + 1) / 2;

  if (x / mid < mid)
    return sqrt_helper<T>(x, lo, mid - 1);
  else
    return sqrt_helper(x, mid, hi);
}

template <typename T>
constexpr T ct_sqrt(T x)
{
  return sqrt_helper<T>(x, 0, x / 2 + 1);
}

Answer 5

If we look at the closest draft standard to C++11 N3337 we can see that sqrt is not marked constexpr, from section 26.8 c.math:

The contents of these headers are the same as the Standard C library headers and respectively, with the following changes:

none of the changes include adding constexpr to sqrt.

We can see from the question Is gcc considering builtins of non-constant expression functions to be constant expressions, that gcc marks many math functions as constexpr as an extension. This extension is a non-conforming extension, as I note in my answer to the linked question when gcc implemented this it looked like it would be a conforming extension but this changed and gcc is likely to fix this this extension to be conforming.