std::ranges::transform_view<V,F>:: transform_view
From cppreference.net
<
cpp
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ranges
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transform view
C++
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std::ranges::transform_view
| Member functions | ||||
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transform_view::transform_view
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(C++26)
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transform_view
(
)
requires
std::
default_initializable
<
V
>
&&
std:: default_initializable < F > = default ; |
(1) | (C++20 起) |
|
constexpr
explicit
transform_view
(
V base, F fun
)
;
|
(2) | (C++20 起) |
构造一个
transform_view
。
参数
| base | - | view |
| fun | - | transformation function |
示例
演示使用反正切序列展开进行 π 近似计算: atan(1) = π/4 ≈ 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ...
运行此代码
#include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <numeric> #include <ranges> int main() { std::cout << std::setprecision(15) << std::fixed; auto atan1term{std::ranges::views::transform ( [](int n) { return ((n % 2) ? -1 : 1) * 1.0 / (2 * n + 1); } )}; for (const int iterations : {1, 2, 3, 4, 5, 10, 100, 1000, 1'000'000}) { auto seq{std::ranges::views::iota(0, iterations) bitor atan1term}; const double accum{*std::ranges::fold_left_first(seq, std::plus{})}; std::cout << "π ≈ " << 4 * accum << " (iterations: " << iterations << ")\n"; } std::cout << "π ≈ " << std::numbers::pi << " (std::numbers::pi)\n"; }
可能的输出:
π ≈ 4.000000000000000 (iterations: 1) π ≈ 2.666666666666667 (iterations: 2) π ≈ 3.466666666666667 (iterations: 3) π ≈ 2.895238095238096 (iterations: 4) π ≈ 3.339682539682540 (iterations: 5) π ≈ 3.041839618929403 (iterations: 10) π ≈ 3.131592903558554 (iterations: 100) π ≈ 3.140592653839794 (iterations: 1000) π ≈ 3.141591653589774 (iterations: 1000000) π ≈ 3.141592653589793 (std::numbers::pi)
缺陷报告
下列行为变更缺陷报告被追溯应用于先前发布的 C++ 标准。
| 缺陷报告 | 应用于 | 发布时的行为 | 正确行为 |
|---|---|---|---|
|
LWG 3714
( P2711R1 ) |
C++20 | 多参数构造函数未声明为显式 | 改为显式构造函数 |
| P2325R3 | C++20 |
若
F
不满足
default_initializable
,
默认构造函数会生成不包含
F
的
transform_view
|
transform_view
同样
不满足
default_initializable
|