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Consider the following function that implements optimised O(log n) exponentiation by squaring:

#include <cstdint> // uintmax
// log-n optimised integer power function. Computes x to the power of y
constexpr std::uintmax_t pow(std::uintmax_t x, std::uintmax_t y) {
    // base cases for efficiency and guaranteed termination
    switch (y) {
    case 0:
        return 1;
    case 1:
        return x;
    case 2:
        return x * x;
    case 3:
        return x * x * x;
    }
    // OTHERWISE:
    std::uintmax_t square_root = pow(x, y / 2);
    // otherwise, work out if y is a multiple of 2 or not
    if (y % 2 == 0) {
        return square_root * square_root;
    } else {
        return square_root * square_root * x;
    }
}

The base cases switch could be rewritten to take advantage of deliberate fallthrough between the various cases to "aggregate" the exponentiation of x from the 0th up to the 3rd power:

// log-n optimised integer power function. Computes x to the power of y
constexpr std::uintmax_t pow(std::uintmax_t x, std::uintmax_t y) {
    // base cases for efficiency and guaranteed termination
    std::uintmax_t result = 1;
    switch (y) {
    case 3:
        result *= x;
        [[fallthrough]];
    case 2:
        result *= x;
        [[fallthrough]];
    case 1:
        result *= x;
        [[fallthrough]];
    case 0:
        return result;
    }
    // OTHERWISE:
    std::uintmax_t square_root = pow(x, y / 2);
    // otherwise, work out if y is a multiple of 2 or not
    if (y % 2 == 0) {
        return square_root * square_root;
    } else {
        return square_root * square_root * x;
    }
}

Now, I am aware that the general consensus is that such use of fallthrough in a switch-case is seen as a cardinal sin. I am also aware that the function would work if base cases were provided for just the zeroth and first power (or even just for the zeroth!). The reason I provide base cases up to the third is that it feels just a bit of a waste of a function call to recur into to calculate such simple powers, although perhaps to the third power is a bit overkill..?

So my question is primarily:

  • Is such use of deliberate fallthrough justifiable here, or could it be justifiable for a similarly-structured example that's a bit more elaborate than chaining multiplication but which still has the commutative chaining as a property?

Secondarily:

  • What do you think about the number of base cases provided here? Is this astute use of deliberately avoiding recursive calls for such simple cases, or is it premature optimisation?
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2 Answers 2

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Firstly, I wouldn't call the function pow - that's confusingly similar to std::pow, which can be invoked with the same arguments.

In both cases, I expect the usual (iterative) binary exponentiation to be both faster and easier for readers to follow than this recursive implementation. Given that the question tags mention performance is important, I encourage you to benchmark both recursive and iterative functions.


Here's a simple implementation of the iterative method:

#include <concepts>

template<typename T>
constexpr T unit_value = 1;


// Return xⁿ
// If calculation overflows, behaviour may be undefined!
template<typename T>
constexpr auto ipow(const T x, std::unsigned_integral auto n)
    requires requires(T t) { t *= t; }
{
    auto result = unit_value<T>;
    for (auto y = x;  true;  y *= y) {
        if (n % 2 == 1) { result *= y; }
        if ((n /= 2) == 0) { break; }
    }
    return result;
}

The infinite loop with break is used rather than testing n > 0 in the for condition to avoid an unnecessary y *= y in the last iteration.

I provided the unit_value template so that it can be used with non-arithmetic types (complex numbers, square matrices, etc) by specialising that value with the appropriate multiplicative identity. For example:

#include <complex>
template<typename T>
constexpr std::complex<T> unit_value<std::complex<T>> = {1, 0};

(This one isn't strictly needed, since there's implicit conversion from T to complex<T>, but it demonstrates the point).

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  • \$\begingroup\$ Thanks, I like this implementation better than the two I offered. Shorter and easier to follow. The unit_value is a neat trick too —I've sometimes needed an overloadable "0" or "1" value for arbitrary numeric-like types before. I can't believe I missed the trick of doing a for-loop with division for the loop-advance clause! \$\endgroup\$
    – saxbophone
    Commented Nov 11, 2022 at 9:53
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    \$\begingroup\$ I've been a bit unconventional with the for loop by declaring y but using n as control variable - some reviewers will object to that, but it's easily changed to an equivalent that should compile to the same object code. \$\endgroup\$ Commented Nov 11, 2022 at 10:34
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    \$\begingroup\$ Nice use of std::unsigned_integral too btw. I like using concepts but forgot they can be used in this way also. \$\endgroup\$
    – saxbophone
    Commented Nov 11, 2022 at 13:56
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    \$\begingroup\$ Good point about break before squaring y - this is one case where the loop fits shell syntax better than C++ (because shell permits multiple commands in both the condition and the body). \$\endgroup\$ Commented Nov 15, 2022 at 7:54
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    \$\begingroup\$ @Matthieu, I updated to avoid the final multiplication (I'm not sure whether a compiler could optimise that away; I suspect it depends on how *= is defined for type T, and whether it can prove there are no side-effects). \$\endgroup\$ Commented Nov 15, 2022 at 8:02
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I would indeed not use [[fallthrough]] here, the version without is clearer in my opinion.

Instead though I would move everything inside the switch-statement:

switch (y) {
case 0:
    return 1;
case 1:
    return x;
case 2:
    return x * x;
case 3:
    return x * x * x;
default:
    return pow(x, y / 2) * pow(x, y - y / 2);
}

Also consider making it a template, possibly in combination with concepts to restrict the type of the exponent to unsigned integers:

template <typename T, std::unsigned_integral Exponent>
constexpr T pow(T x, Exponent y) {
    ...
}
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    \$\begingroup\$ Thanks, that's a useful point about putting the whole thing in a switch. Does the implementation in the default case that you provide still exhibit O(log n) performance? I was under the impression that one can implement with fewer recursive calls if one computes pow(x, y / 2) into a temporary and computes the square of this, multiplied by the remaining product (if any) when odd. I can always make the default a braced-case I suppose \$\endgroup\$
    – saxbophone
    Commented Nov 11, 2022 at 9:11
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    \$\begingroup\$ You're right, that might not be the case. I think @TobySpeight's answer is even better, it avoids recursion altogether. \$\endgroup\$
    – G. Sliepen
    Commented Nov 11, 2022 at 13:19

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