问题描述
标题是主要问题.确切的情况(我正在使用命名空间std;"):
The title is the main question. The exact scenario (I am 'using namespace std;'):
void SubstringMiner::sortByOccurrence(list<Substring *> & substring_list) {
list::sort(substring_list.begin(), substring_list.end(), Substring::OccurrenceComparator);
}
这是比较器的定义:
class Substring {
// ...
class OccurrenceComparator {
public:
bool operator() (Substring * a, Substring *b);
}
};
比较器的实现是直观而琐碎的.我还在std :: set中使用了非常类似的比较器,并且工作正常.当我添加sortByOccurrence()函数时,会出现标题错误.
Implementation of the comparator is intuitive and trivial. I am also using a very similar comparator in a std::set and it works fine. When I add the sortByOccurrence() funcion it gives me the error in the title.
我该怎么办?
编辑:我现在正尝试将Substring :: OccurrenceComparator()用作比较器,并收到以下错误消息:
I'm now trying to pass Substring::OccurrenceComparator() as the comparator, and am getting the following error:
g++ -Wall -g -c substring_miner.cpp -o obj/subtring_miner.o
substring_miner.cpp: In function ‘void SubstringMiner::sortByOccurrence(std::list<Substring*, std::allocator<Substring*> >&)’:
substring_miner.cpp:113: error: no matching function for call to ‘std::list<Substring*, std::allocator<Substring*> >::sort(std::_List_iterator<Substring*>, std::_List_iterator<Substring*>, Substring::OccurrenceComparator)’
/usr/include/c++/4.3/bits/list.tcc:303: note: candidates are: void std::list<_Tp, _Alloc>::sort() [with _Tp = Substring*, _Alloc = std::allocator<Substring*>]
make: *** [substring_miner] Error 1
我的代码行现在是:
list<Substring *>::sort(substring_list.begin(), substring_list.end(), Substring::OccurrenceComparator());
我无法删除模板,否则给我一个错误,提示模板参数错误.
I can't remove the template or it gives me an error saying that template parameters were wrong.
推荐答案
list
成员sort
是非静态函数,因此必须在列表实例上调用.
list
member sort
is a non-static function so must be called on a list instance.
substring_list.sort( Substring::OccurrenceComparator() );
编辑:您不能使用免费功能std::sort
,因为它需要随机访问迭代器,而list
迭代器则不需要.
You can't use the free function std::sort
as it requires random access iterators which list
iterators are not.
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