C++20：如何按大小分割范围？

Question

我想将范围 {1, 2, 3, 4, 5} 拆分为 <任意大小> 的子范围范围（例如，大小为 2：{{1, 2}, {3, 4}, {5}}）。然而 std::views::split 仅按分隔符拆分。

没有标准的“反向连接”或其他方法吗？

Answer 1

range-v3 调用此算法 chunk. There is no such range adapter in C++20, but it is part of the set being proposed for C++23 under the same name. For example:

#include <vector>
#include <range/v3/view/chunk.hpp>
#include <fmt/format.h>
#include <fmt/ranges.h>

int main() {
    std::vector v = {1, 2, 3, 4, 5};
    fmt::print("{}\n", v | ranges::views::chunk(2)); // prints {{1, 2}, {3, 4}, {5}}
}

这似乎是跨语言为该算法选择的非常一致的名称。 Python有chunked, Rust has chunks, Swift has chunks(ofCount: n), D has chunks等

C++20：如何按大小分割范围？

C++20: How to split range by size?

c++

range

c++20

std-ranges