从 STL 集中删除最大元素

发布于 2024-09-11 15:48:42 字数 614 浏览 1 评论 0原文

这是我之前提出的问题的后续问题（STL max_element的复杂性）。

我基本上想从集合中弹出最大元素，但我遇到了问题。

大致是我的代码：

set<Object> objectSet;

Object pop_max_element() {
    Object obj = *objectSet.rbegin();
    set<Object>::iterator i = objectSet.end()--; //this seems terrible
    objectSet.erase(i); //*** glibc detected *** free(): invalid pointer
    return obj;
}

早些时候我尝试过 objectSet.erase(objectSet.rbegin()); 但编译器抱怨没有匹配的函数（我猜它不喜欢反向迭代器）。我知道没有检查空集，但当 objectSet.size() >> 时失败。 0。

原文

This is a follow-up on a previous question I had ( Complexity of STL max_element ).

I want to basically pop the max element from a set, but I am running into problems.

Here is roughly my code:

set<Object> objectSet;

Object pop_max_element() {
    Object obj = *objectSet.rbegin();
    set<Object>::iterator i = objectSet.end()--; //this seems terrible
    objectSet.erase(i); //*** glibc detected *** free(): invalid pointer
    return obj;
}

Earlier I tried objectSet.erase(objectSet.rbegin()); but the compiler complained that there was no matching function (I'm guessing it doesn't like the reverse_iterator). I know there is no checking for an empty set, but it's failing when objectSet.size() >> 0.

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不忘初心 2024-09-18 15:48:42

你已经非常接近了，但是你试图在迭代器分配中做太多的事情。您将后递减运算符应用于end返回的任何内容。我不太确定那是做什么的，但几乎可以肯定这不是你想要的。将 end 的结果赋给 i，然后然后将其递减以获得集合的最后一个元素。

set<Object>::iterator i = objectSet.end();
--i;
Object obj = *i;
objectSet.erase(i);
return obj;

You're pretty close, but you're trying to do a little too much in that iterator assignment. You're applying the post-decrement operator to whatever end returns. I'm not really sure what that does, but it's almost certainly not what you want. Assign the result of end to i, and then decrement it to get the last element of the set.

set<Object>::iterator i = objectSet.end();
--i;
Object obj = *i;
objectSet.erase(i);
return obj;

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你不是我要的菜∠ 2024-09-18 15:48:42

你需要这样做：

set<Object> objectSet;

Object pop_max_element() {
    Object obj = *objectSet.rbegin();
    set<Object>::iterator i = --objectSet.end(); // NOTE: Predecrement; not postdecrement.
    objectSet.erase(i); //*** glibc detected *** free(): invalid pointer
    return obj;
}

You need to do this:

set<Object> objectSet;

Object pop_max_element() {
    Object obj = *objectSet.rbegin();
    set<Object>::iterator i = --objectSet.end(); // NOTE: Predecrement; not postdecrement.
    objectSet.erase(i); //*** glibc detected *** free(): invalid pointer
    return obj;
}

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雨夜星沙 2024-09-18 15:48:42

该语句的

set<Object>::iterator i = objectSet.end()--;

意思是“将 end() 分配给 i，然后递减一个即将被丢弃的临时变量”。换句话说，它与 set

assert(!objectSet.empty()); // check there is something before end
set<Object>::iterator i = objectSet.end();
--i;
objectSet.erase(i);

没关系，这是本质上为集合重现 .back() 的合法方法。

另外，反向迭代器有一个 base() 成员可以转换为普通迭代器，我猜你只能删除普通迭代器 - 尝试 objectSet.erase(objectSet.rbegin().base() ）。

The statement

set<Object>::iterator i = objectSet.end()--;

means 'assign end() to i then decrement a temporary variable that is about to be thrown away'. In other words it's the same as set<Object>::iterator i = objectSet.end();, and I'm sure you recognise you cannot erase end(), because it points to one past the end. Use something like this instead:

assert(!objectSet.empty()); // check there is something before end
set<Object>::iterator i = objectSet.end();
--i;
objectSet.erase(i);

and that's okay, it's a legitimate way to essentially reproduce .back() for a set.

Also, reverse iterators have a base() member to convert to a normal iterator and I guess you can only erase normal iterators - try objectSet.erase(objectSet.rbegin().base()).

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