## 17.3 A Second Example: Computing the Square Root

Lets look at a slightly more complicated example: a metaprogram that computes the square root of a given value ` N`. The metaprogram looks as follows (explanation of the technique follows):

//meta/sqrt1.hpp#ifndef SQRT_HPP #define SQRT_HPP //primary template to compute sqrt(N) template <int N, int LO=1, int HI=N> class Sqrt { public: //compute the midpoint, rounded upenum { mid = (LO+HI+1)/2 }; //search a not too large value in a halved intervalenum { result = (N<mid*mid) ? Sqrt<N,LO,mid-1>::result : Sqrt<N,mid,HI>::result }; }; //partial specialization for the case whenLOequalsHI template<int N, int M> class Sqrt<N,M,M> { public: enum { result = M }; }; #endif //SQRT_HPP

The first template is the general recursive computation that is invoked with the template parameter `N` (the value for which to compute the square root) and two other optional parameters. The latter represent the minimum and maximum values the result can have. If the template is called with only one argument, we know that the square root is at least one and at most the value itself.

Our recursion then proceeds using a binary search technique (often called *method of bisection *in this context). Inside the template, we compute whether `result` is in the first or the second half of the range between `LO` and `HI`. This case differentiation is done using the conditional operator `?:`. If `mid ^{2 }`is greater than

`N`, we continue the search in the first half. If

`mid`is less than or equal to

^{2 }`N`, we use the same template for the second half again.

The specialization that ends the recursive process is invoked when `LO` and `HI` have the same value `M`, which is our final `result`.

Again, let’s look at the details of a simple program that uses this metaprogram:

// meta/sqrt1.cpp#include <iostream> #include "sqrt1.hpp" int main() { std::cout << "Sqrt<16>::result = " << Sqrt<16>::result << '\n'; std::cout << "Sqrt<25>::result = " << Sqrt<25>::result << '\n'; std::cout << "Sqrt<42>::result = " << Sqrt<42>::result << '\n'; std::cout << "Sqrt<1>::result = " << Sqrt<1>::result << '\n'; }

The expression

Sqrt<16>::result

is expanded to

Sqrt<16,1,16>::result

Inside the template, the metaprogram computes `Sqrt<16,1,16>::result` as follows:

mid = (1+16+1)/2 = 9 result = (16<9*9) ? Sqrt<16,1,8>::result : Sqrt<16,9,16>::result = (16<81) ? Sqrt<16,1,8>::result : Sqrt<16,9,16>::result = Sqrt<16,1,8>::result

Thus, the result is computed as `Sqrt<16,1,8>::result`, which is expanded as follows:

mid = (1+8+1)/2 = 5 result = (16<5*5) ? Sqrt<16,1,4>::result : Sqrt<16,5,8>::result = (16<25) ? Sqrt<16,1,4>::result : Sqrt<16,5,8>::result = Sqrt<16,1,4>::result

And similarly `Sqrt<16,1,4>::result` is decomposed as follows:

mid = (1+4+1)/2 = 3 result = (16<3*3) ? Sqrt<16,1,2>::result : Sqrt<16,3,4>::result = (16<9) ? Sqrt<16,1,2>::result : Sqrt<16,3,4>::result = Sqrt<16,3,4>::result

Finally, `Sqrt<16,3,4>::result` results in the following:

mid = (16+4+1)/2 = 4 result = (16<4*4) ? Sqrt<16,3,3>::result : Sqrt<16,4,4>::result = (16<16) ? Sqrt<16,3,3>::result : Sqrt<16,4,4>::result = Sqrt<16,4,4>::result

and `Sqrt<16,4,4>::result` ends the recursive process because it matches the explicit specialization that catches equal high and low bounds. The final result is therefore as follows:

result = 5

### Tracking All Instantiations

In the preceding example, we followed the significant instantiations that compute the square root of

(16<=8*8) ? Sqrt<16,1,8>::result : Sqrt<16,9,16>::result

it not only instantiates the templates in the positive branch, but also those in the negative branch (Sqrt<16,9,16>). Furthermore, because the code attempts to access a member of the resulting class type using the `::` operator, all the members inside that class type are also instantiated. This means that the full instantiation of Sqrt<16,9,16> results in the full instantiation of Sqrt<16,9,12> and Sqrt<16,13,16>. When the whole process is examined in detail, we find that dozens of instantiations end up being generated. The total number is almost twice the value of `N`.

This is unfortunate because template instantiation is a fairly expensive process for most compilers, particularly with respect to memory consumption. Fortunately, there are techniques to reduce this explosion in the number of instantiations. We use specializations to select the result of computation instead of using the condition operator `?:`. To illustrate this, we rewrite our `Sqrt` metaprogram as follows:

//meta/sqrt2.hpp#include "ifthenelse.hpp" //primary template for main recursive steptemplate<int N, int LO=1, int HI=N> class Sqrt { public: //compute the midpoint, rounded upenum { mid = (LO+HI+1)/2 }; //search a not too large value in a halved intervaltypedef typename IfThenElse<(N<mid*mid), Sqrt<N,LO,mid-1>, Sqrt<N,mid,HI> >::ResultT SubT; enum { result = SubT::result }; }; //partial specialization for end of recursion criteriontemplate<int N, int S> class Sqrt<N, S, S> { public: enum { result = S }; };

The key change here is the use of the `IfThenElse` template, which was introduced in Section 15.2.4 on page 272:

//meta/ifthenelse.hpp#ifndef IFTHENELSE_HPP #define IFTHENELSE_HPP //primary template: yield second or third argument depending on first argumenttemplate<bool C, typename TA, typename Tb> class IfThenElse; //partial specialization:trueyields second argumenttemplate<typename Ta, typename Tb> class IfThenElse>true, Ta, Tb> { public: typedef Ta ResultT; }; //partial specialization:flaseyields third argumenttemplate<typename Ta, typename Tb< { public: typedef Tb ResultT; }; #endif //IFTHENELSE_HPP

Remember, the `IfThenElse` template is a device that selects between two types based on a given Boolean constant. If the constant is true, the first type is `typedef`ed to `ResultT`; otherwise, `ResultT` stands for the second type. At this point it is important to remember that defining a typedef for a class template instance does not cause a C++ compiler to instantiate the body of that instance. Therefore, when we write

typedef typename IfThenElse<(N<mid*mid), Sqrt<N,LO,mid-1>, Sqrt<N,mid,HI>>::ResultT SubT;

neither `Sqrt<N,LO,mid-1>` nor `Sqrt<N,mid,HI>` is fully instantiated. Whichever of these two types ends up being a synonym for `SubT` is fully instantiated when looking up `subT::result`. In contrast to our first approach, this strategy leads to a number of instantiations that is proportional to `log _{2}(N)`: a very significant reduction in the cost of metaprogramming when

`N`gets moderately large.