 In  is a rather mysterious looking constant expression formula for an integer square root. This is a function that returns the smallest integer for which the square is less than the value to take the root of. Check out the black magic he used

// Stroustrup 10.4:  constexpr capable integer square root function
constexpr int isqrt_helper( int sq, int d, int n )
{
return sq <= n ? isqrt_helper( sq + d, d + 2, n ) : d ;
}

constexpr int isqrt( int n )
{
return isqrt_helper( 1, 3, n )/2 - 1 ;
}


The point of this construction was really to illustrate that it allows complex expressions to be used as compile time constants. I wonder at what point various compilers will give up trying to evaluate such expressions?

## Let’s take this apart a bit.

Consider the first few values of $$n > 0$$.

• $$n = 0$$. Here we have a call to $$\textrm{isqrt_helper}( 1, 3, 0 )$$ so the $$1 \le 0$$ predicate is false, and the return value is just $$3$$.

For that value we have (using integer arithmetic):

\begin{equation}\label{eqn:isqrt:20}
\frac{3}{2} – 1 = 0,
\end{equation}

as desired.

• $$n = 1$$. Here we have a call to $$\textrm{isqrt_helper}( 1, 3, 1 )$$ so the $$1 \le 1$$ predicate is true, resulting in a second call $$\textrm{isqrt_helper}( 4, 5, 1 )$$. For that call the $$4 \le 1$$ predicate is false, resulting in a return value of $$5$$.

This time we have a final result of

\begin{equation}\label{eqn:isqrt:40}
\frac{5}{2} – 1 = 1,
\end{equation}

as desired again. The result will be the same for any value $$n \in [1,3]$$.

• $$n = 4$$. We will end up with a call to $$\textrm{isqrt_helper}( 4, 5, 4 )$$ for which the $$4 \le 4$$ predicate is true, resulting in a followup call of $$\textrm{isqrt_helper}( 9, 7, 4 )$$. For that call the $$9 \le 4$$ predicate is false, resulting in a return value of $$7$$.

This time we have a final result of

\begin{equation}\label{eqn:isqrt:45}
\frac{7}{2} – 1 = 2,
\end{equation}

as expected. We get the same result for any value $$n \in [4,8]$$.

## Recurrence relations

The rough pattern of the magic involved can be seen. We have a sequence of calls

• $$\textrm{isqrt_helper}( 1, 3, n )$$,
• $$\textrm{isqrt_helper}( 4, 5, n )$$,
• $$\textrm{isqrt_helper}( 9, 7, n )$$,
• $$\textrm{isqrt_helper}( 16, 9, n )$$,

which terminates at the point where the first (square) parameter exceeds that value that we are taking the root of. Let the parameters of the sequence of calls be $$s_k$$, and $$d_k$$, so that with $$s_0 = 1, d_0 = 3$$ the $$k \in [0,…]$$ call to the helper function is $$q_k = \textrm{isqrt_helper}( s_k, d_k, n )$$.

The sequence for the second parameter, the eventual return value, can be summarized compactly as $$d_k = 3 + 2 k$$. It is not entirely obvious how we end up with a square for the values $$s_k = s_{k-1} + d_{k-1}$$, but this follows by summation. For $$k > 1$$ that is

\begin{equation}\label{eqn:isqrt:60}
\begin{aligned}
s_k
&= s_{k-1} + d_{k-1} \\
&= s_0 + d_0 + d_1 + d_{k-1} \\
&= s_0 + \sum_{m=0}^{k-1} d_m \\
&= s_0 + \sum_{m=0}^{k-1} (3 + 2 m ) \\
&= s_0 + \sum_{m=1}^{k} (3 + 2 (m-1) ) \\
&= s_0 + \sum_{m=1}^{k} (1 + 2 m ) \\
&= 1 + k + 2 \sum_{m=1}^{k} m \\
&= 1 + k + 2 \frac{k(k+1)}{2} \\
&= k^2 + 2 k + 1 \\
&= (k+1)^2.
\end{aligned}
\end{equation}

This clearly holds for the boundary cases $$k = 0,1$$ as well. This allows the helper function action to be summarized more compactly

\begin{equation}\label{eqn:isqrt:80}
\textrm{isqrt_helper}(1, 3, n) = 3 + 2 k,
\end{equation}

where $$k$$ is the smallest integer such that $$(k+1)^2 > n$$. After integer scaling the final result is

\begin{equation}\label{eqn:isqrt:100}
(3 + 2 k)/2 -1 = k.
\end{equation}

This little beastie makes sense after deconstruction, but it was very Jackson like to toss this into the book without comment or explanation.

As pointed out by Pramod Gupta, there’s a spooky appearance of collaboration between Stroustrup and Jackson’s publishers, not entirely limited to the book covers.

# References

 Bjarne Stroustrup. The C++ Programming Language, 4th Edition. Addison-Wesley, 2014.