从指针获取迭代器 [英] Get iterator from pointer

问题描述

以下问题细分错误执行方法我在另一个问题中停止了。

Following question Segmentation fault executing method I stopped in another problem.

给定：

Cluster * base;
Cluster * absorbed;

list<Cluster>::iterator li = clusters.begin();

// li is then pointed to some element.

absorbed = &(*li);   // Get a pointer to the chosen element.

base->joinCluster(absorbed);   // Perform an action with absorbed.

// li contines the cycle and now point to clusters.end()

// Now, absorbed should be deleted from list. li doesn't point to absorbed anymore.
li = & (* absorbed);
clusters.erase(li);

尝试执行上述相反操作，但g ++返回错误：错误：不匹配'operator ='in'li = absorb'
我该如何做？

Tried to do the reverse of above, but g++ returns an error: error: no match for 'operator=' in 'li = absorbed' How can I do this?

避免循环遍历列表中的所有元素以再次找到吸收。

I'm trying to avoid to cycle through all the elements in the list to find again the absorbed one.

编辑：

很抱歉，我理解我错过了 absorb =&（* li）; li 继续循环，直到 clusters.end（）。
因此，在循环结束时， li 不再指向吸收。
我想我有两个选项：我要么循环再次通过集群中的所有元素找到吸收元件;或者我利用吸收的指针来从集群中擦除元素。但是，我如何实现这第二个选项？

Sorry, but i understand I missed to say that after absorbed = &(* li); li continues to cycle until clusters.end(). So, at the time the cycle ends, li doesn't point to absorbed anymore. I think I have two option: Either I cycle again through all elements in clusters to find the absorbed element; Or I take advantage of the absorbed pointer to erase the element from clusters. But, how can I achieve this second option?

推荐答案

跳过 li =&（* absorb）它应该工作很好。

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