递归地传递一个字符串而不需要娱乐

Pass a string Recursively without Recreation(递归地传递一个字符串而不需要娱乐)

问题描述

我在这里回答了一个问题:https://stackoverflow.com/a/28862668/2642059我需要在哪里使用重复来遍历 string.我想在每个函数上使用 const string& 作为我的参数，但是除非我想在每次递归时重建字符串，否则我发现我需要传递一个 start 和 finish 位置以及 string 本身.所以传递 string 变得毫无意义.

I answered a question here: https://stackoverflow.com/a/28862668/2642059 Where I needed to use recurrence to step through a string. I wanted to use a const string& as my parameter on each function, but unless I wanted to reconstruct the string each recursion I found that I needed to pass a start and finish position as well as the string itself. So it became pointless to pass the string at all.

最后我选择只将 start 和 finish 指针传递给 char[].

In the end I choose to just pass a start and finish pointer to the char[].

举个例子，假设我得到一个包含嵌套括号的字符串(但没有并排插入括号.)就像这样:

As an example, say that I'm given a string which contains nested parenthesis (but no side by side parenthetical insertions.) So like this:

(abc(def(ghi((j)klm)nop)qrs)tuv)wxyz

(abc(def(ghi((j)klm)nop)qrs)tuv)wxyz

但不是这样:

(abc(def)(ghi)(j)(klm)(nop)(qrs)tuv)wxyz

(abc(def)(ghi)(j)(klm)(nop)(qrs)tuv)wxyz

我想编写一个递归程序来提取最深嵌套括号中的字符串.类似的东西:

I want to write a recursive program to extract the string in the deepest nested parentheses. Something like:

string foo(const string& bar){
    auto start = bar.find('(') + 1;

    return start == string::npos + 1 ? bar : foo(bar.substr(start, bar.find_last_of(')') - start));
}

然而，我不高兴为 foo 的每次重复重建一个 string.另一种方法是像链接示例中一样传递 start 和 finish 指针(或传递 string::const_iterators.)

However I'm unhappy reconstructing a string for each recurrence of foo. The alternative is to pass start and finish pointers as in the linked example (or to pass string::const_iterators.)

是否有包装器或其他东西可以让我使用 string 功能，但不能重建 string?

Is there a wrapper or something which would allow me to use string functionality, but not reconstruct a string?

推荐答案

string_view 来自库基础知识 TS 可能是一个想法，在 GCC 中提供支持.

string_view from the library fundamentals TS might be one idea, support is available in GCC.

接口实际上与string

#include <experimental/string_view>
using std::experimental::string_view;

string_view foo(const string_view& bar){
    auto start = bar.find('(') + 1;

    return start == string_view::npos + 1 ? bar : foo(bar.substr(start, bar.find_last_of(')') - start));
}

最后一行也可以

return start ? foo(bar.substr(start, bar.find_last_of(')') - start)) : bar;

虽然它们都很神秘.

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