問題描述

我在這里回答了一個問題:https://stackoverflow.com/a/28862668/2642059我需要在哪里使用重復(fù)來遍歷 string.我想在每個函數(shù)上使用 const string& 作為我的參數(shù)，但是除非我想在每次遞歸時重建字符串，否則我發(fā)現(xiàn)我需要傳遞一個 start 和 finish 位置以及 string 本身.所以傳遞 string 變得毫無意義.

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.

最后我選擇只將 start 和 finish 指針傳遞給 char[].

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

舉個例子，假設(shè)我得到一個包含嵌套括號的字符串(但沒有并排插入括號.)就像這樣:

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

但不是這樣:

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

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

我想編寫一個遞歸程序來提取最深嵌套括號中的字符串.類似的東西:

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，我不高興為 foo 的每次重復(fù)重建一個 string.另一種方法是像鏈接示例中一樣傳遞 start 和 finish 指針(或傳遞 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.)

是否有包裝器或其他東西可以讓我使用 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?

推薦答案

string_view 來自庫基礎(chǔ)知識 TS 可能是一個想法，在 GCC 中提供支持.

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

接口實際上與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;

雖然它們都很神秘.

這篇關(guān)于遞歸地傳遞一個字符串而不需要娛樂的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！