Lazy变换在C ++ [英] Lazy transform in C++

问题描述

我有以下的Python代码片段，我想使用C ++重现：

I have the following Python snippet that I would like to reproduce using C++:

from itertools import count, imap

source = count(1)
pipe1 = imap(lambda x: 2 * x, source)
pipe2 = imap(lambda x: x + 1, pipe1)
sink = imap(lambda x: 3 * x, pipe2)
for i in sink:
    print i

我听说过 Boost Phoenix a>，但我找不到一个懒惰的例子 transform 的行为与Python的 imap 相同。

I've heard of Boost Phoenix, but I couldn't find an example of a lazy transform behaving in the same way as Python's imap.

编辑：为了澄清我的问题，这个想法不仅是使用为按顺序应用函数，而是能够在无限生成器上使用 std :: transform 的算法。

to clarify my question, the idea is not only to apply functions in sequence using a for, but rather to be able to use algorithms like std::transform on infinite generators. The way the functions are composed (in a more functional language like dialect) is also important, as the next step is function composition.

更新：感谢bradgonesurfing，David Brown，和Xeo的惊人答案！我选择了Xeo's，因为它是最简洁的，它让我在我想成为的地方，但大卫是非常重要的通过的概念。此外，bradgonesurfing的提示Boost :: Range：）。

Update: thanks bradgonesurfing, David Brown, and Xeo for the amazing answers! I chose Xeo's because it's the most concise and it gets me right where I wanted to be, but David's was very important into getting the concepts through. Also, bradgonesurfing's tipped Boost::Range :).

推荐答案

雇用 Boost.Range ：

int main(){
  auto map = boost::adaptors::transformed; // shorten the name
  auto sink = generate(1) | map([](int x){ return 2*x; })
                          | map([](int x){ return x+1; })
                          | map([](int x){ return 3*x; });
  for(auto i : sink)
    std::cout << i << "\n";
}

Live示例，包括生成函数

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