ChrisIzatt - 1 year ago 102
C++ Question

# How do I express this?

I'm trying to work out how to write the following:

``````total = (value * 0.95 ^ 0) + (value * 0.95 ^ 1) + (value * 0.95 ^ 2) ...
``````

or:

``````x = (y * z ^ 0) + (y * z ^ 1) + (y * z ^ 2) + (y * z ^ 3) ...
``````

This expresses how to calculate x for 4 iterations, but how can I express this to work with a variable number of iterations? Obviously I could create a loop and add the values together, but I'd really like to find a single equation that solves this.

I'm using c++ but I guess this isn't really a language specific problem (sorry I literally don't know where else to ask this question!).

Any ideas?

Thanks,
Chris.

Equation-wise solving, this is a geometric series and can therefore be calculated with

``````double geometric_series(double y, double z, int N) {
return y * (std::pow(z, N) - 1.0) / (z - 1.0);
}
``````

but the same result can be obtained with some fun C++ metaprogramming: if you know the number of iterations in advanced and you're allowed to use C++17 features and fold expressions you could do as follows

``````template<std::size_t... N>
double calculate_x(double y, double z, std::index_sequence<N...>) { // [0;N[
auto f = [](double y_p, double z_p, double exp) {
return y_p * std::pow(z_p, exp);
};
return (f(y, z, N) + ...);
}

template <std::size_t N>
auto calculate_x(double y, double z) {
return calculate_x(y, z, std::make_index_sequence<N>{});
}
``````

Alternatively this can also be done with pre-C++17 templates

``````template <int N>
double calculate_x(double y, double z) {
return calculate_x<N-1>(y, z) + (y * std::pow(z, N - 1));
}

template <>
double calculate_x<0>(double, double) {
return 0;
}
``````

Otherwise a simpler solution would be to just use a loop

``````double calculate_x_simple(double y, double z, int N) {
double ret = 0.0;
for (int i = 0 ; i < N ; ++i)
ret += y * std::pow(z, i);
return ret;
}
``````

Driver for the code above

``````int main() {

// x = (y * z ^ 0) + (y * z ^ 1) + (y * z ^ 2) + (y * z ^ 3)
double y = 42.0;
double z = 44.5;
std::cout << (calculate_x<3>(y, z) == calculate_x_simple(y, z, 3)); // 1

}
``````
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