现代C ++：初始化constexpr表 [英] Modern C++: initialize constexpr tables

问题描述

假设我有一个类 X ，该功能需要大量常量表值，例如数组 A [1024] 。我有一个递归函数 f 计算其值，类似

Suppose I have a class X, which functionality requires a lot of constant table values, say an array A[1024]. I have a recurrent function f that computes its values, smth like

A[x] = f(A[x - 1]);

假设 A [0] 是一个已知常数，因此数组的其余部分也是常数。使用现代C ++的功能并且不使用此数组的硬编码值存储文件的情况下，预先计算这些值的最佳方法是什么？我的解决方法是const静态伪变量：

Suppose that A[0] is a known constant, therefore the rest of the array is constant too. What is the best way to calculate these values beforehand, using features of modern C++, and without storaging file with hardcoded values of this array? My workaround was a const static dummy variable:

const bool X::dummy = X::SetupTables();

bool X::SetupTables() {
    A[0] = 1;
    for (size_t i = 1; i <= A.size(); ++i)
        A[i] = f(A[i - 1]);
}

但是我相信，这并不是最美丽的方式。
注意：我强调数组很大，我想避免代码的怪异。

But I believe, it’s not the most beautiful way to go. Note: I emphasize that array is rather big and I want to avoid monstrosity of the code.

推荐答案

自C ++ 14起， for 循环就可以在 constexpr 函数。此外，自C ++ 17起， std :: array： ：operator [] 也是 constexpr 。

Since C++14, for loops are allowed in constexpr functions. Moreover, since C++17, std::array::operator[] is constexpr too.

因此，您可以写这样的东西：

So you can write something like this:

template<class T, size_t N, class F>
constexpr auto make_table(F func, T first)
{
    std::array<T, N> a {first};
    for (size_t i = 1; i < N; ++i)
    {
        a[i] = func(a[i - 1]);
    }
    return a;
}

示例： https://godbolt.org/g/irrfr2

这篇关于现代C ++：初始化constexpr表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！