如何使用std :: forward_list在常量时间范围内进行拼接？

问题描述

我想将 [first，last] 与两个端点（包括端点）进行拼接。我有之前 第一和最后的元素的迭代器。我可以用 splice_after（）但只在线性时间。

/ code>？

如果问题不清楚，这里是显示我的问题的示例代码：

上的代码

  #include< algorithm> 
 #include< forward_list> 
 #include< iostream> 
 #include  using namespace std; 
 
 int main（）{
 forward_list< char> trg {'a'，'b'，'c'} 
 forward_list< char> src {'1'，'2'，'3'，'4'}; 
 
 auto before_first = src.begin（）; 
 auto last = find（src.begin（），src.end（），'4'）; 
 cout<< before_first =<< * before_first<< ，last =< * last<< \\\
; 
 
 // trg.splice（trg.begin（），src，before_first，last）; // no such splice 
 auto end = last; 
 ++ end; //哎哟！ splice已经找到了最后一次，虽然我已经有了：
 trg.splice_after（trg.begin（），src，before_first，end）; 
 
 cout< splice：\\\
; 
 copy（trg.begin（），trg.end（），ostream_iterator< char>（cout，））; 
 
 cout< \\\
Source after splice：\\\
; 
 copy（src.begin（），src.end（），ostream_iterator< char>（cout，））; 
 
 cout<<< endl; 
} 
  

输出：

  before_first = 1，last = 4 
拼接后目标：
a 2 3 4 bc 
源拼接后：
 1 
  

解决方案

规范 forward_list 说，范围（第一个，最后）应该拼接，不幸的是没有办法这在O（1）的时间，因为一个需要访问 last-1 这样做，以及访问 last-1 是从开始向前迭代。

如果规格是拼接范围（first，last] ，那么O（1）拼接将是可能的。我知道没有办法实现这个与当前的 forward_list 规范。

我认为这是一个缺陷。但我已尝试并无法解决问题：

然而，过去的问题已经扭转，特别是当投诉来自非委员会成员，如你自己。提出投诉的方式是开启新问题，在适当情况下引用任何旧的或相关问题。有关打开问题的说明，请点击。

PS：+1的问题。

I want to splice the range [first, last], with both endpoints inclusive. I have iterators to the element before first and to last. I could do it with splice_after() but only in linear time.

I belive this splice can be done in constant time. How can I do it with std::forward_list?

If the question is not clear, here as is an example code showing my problem:

Code on Live Work Space

#include <algorithm>
#include <forward_list>
#include <iostream>
#include <iterator>
using namespace std;

int main() {
    forward_list<char> trg{'a','b','c'};
    forward_list<char> src{'1','2','3','4'};

    auto before_first = src.begin();
    auto last = find(src.begin(), src.end(), '4');
    cout << "before_first = " << *before_first << ", last = " << *last << "\n";

    // trg.splice(trg.begin(), src, before_first, last); // no such splice
    auto end = last;
    ++end; // Ouch! splice has to find last again although I already had it  :(
    trg.splice_after(trg.begin(), src, before_first, end);

    cout << "Target after splice:\n";
    copy(trg.begin(), trg.end(), ostream_iterator<char>(cout," "));

    cout << "\nSource after splice:\n";
    copy(src.begin(), src.end(), ostream_iterator<char>(cout," "));

    cout << endl;
}

Output:

before_first = 1, last = 4
Target after splice:
a 2 3 4 b c
Source after splice:
1

解决方案

The specification of forward_list says that the range (first, last) should be spliced, and there is unfortunately no way to do this in O(1) time because one needs access to last-1 to do that, and the only way to get access to last-1 is to iterate forward from first.

Had the spec been to splice the range (first, last], then an O(1) splice would be possible. I know of no way to achieve this with the current forward_list spec.

I think it is a defect. However I've already tried and failed to fix it:

http://cplusplus.github.com/LWG/lwg-defects.html#897

However issues have been reversed in the past, especially when complaints come in from non-committee members such as yourself. The way to file a complaint is to open a new issue, referencing any old or related issues if appropriate. The instructions for opening an issue are here.

PS: +1 on the question.

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