C++ 给定位置，如何获取列表中的某个元素？
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How to get a certain element in a list, given the position?
提问by Test
So I've got a list:
所以我有一个清单：
list<Object> myList;
myList.push_back(Object myObject);
I'm not sure but I'm confident that this would be the "0th" element in the array. Is there any function I can use that will return "myObject"?
我不确定，但我相信这将是数组中的“第 0”个元素。是否有任何我可以使用的函数可以返回“myObject”？
Object copy = myList.find_element(0);
?
?
回答by James McNellis
If you frequently need to access the Nth element of a sequence, std::list, which is implemented as a doubly linked list, is probably not the right choice. std::vectoror std::dequewould likely be better.
如果您经常需要访问序列的第 N 个元素，那么std::list实现为双向链表的 可能不是正确的选择。 std::vector或者std::deque可能会更好。
That said, you can get an iterator to the Nth element using std::advance:
也就是说，您可以使用以下方法获取第 N 个元素的迭代器std::advance：
std::list<Object> l;
// add elements to list 'l'...
unsigned N = /* index of the element you want to retrieve */;
if (l.size() > N)
{
std::list<Object>::iterator it = l.begin();
std::advance(it, N);
// 'it' points to the element at index 'N'
}
For a container that doesn't provide random access, like std::list, std::advancecalls operator++on the iterator Ntimes. Alternatively, if your Standard Library implementation provides it, you may call std::next:
对于不提供随机访问的容器，例如std::list，std::advance调用operator++迭代器N时间。或者，如果您的标准库实现提供了它，您可以调用std::next：
if (l.size() > N)
{
std::list<Object>::iterator it = std::next(l.begin(), N);
}
std::nextis effectively wraps a call to std::advance, making it easier to advance an iterator Ntimes with fewer lines of code and fewer mutable variables. std::nextwas added in C++11.
std::next有效地包装了对 的调用std::advance，从而可以更轻松地N使用更少的代码行和更少的可变变量来推进迭代器次数。 std::next是在 C++11 中添加的。
回答by Nawaz
std::listdoesn't provide any function to get element given an index. You may try to get it by writing some code, which I wouldn't recommend, because that would be inefficient if you frequently need to do so.
std::list不提供任何函数来获取给定索引的元素。您可以尝试通过编写一些代码来获得它，我不建议这样做，因为如果您经常需要这样做，效率会很低。
What you need is : std::vector. Use it as:
你需要的是：std::vector。将其用作：
std::vector<Object> objects;
objects.push_back(myObject);
Object const & x = objects[0]; //index isn't checked
Object const & y = objects.at(0); //index is checked
回答by furas
std::list<Object> l;
std::list<Object>::iterator ptr;
int i;
for( i = 0 , ptr = l.begin() ; i < N && ptr != l.end() ; i++ , ptr++ );
if( ptr == l.end() ) {
// list too short
} else {
// 'ptr' points to N-th element of list
}
回答by Bill Moore
Maybe not the most efficient way. But you could convert the list into a vector.
也许不是最有效的方式。但是您可以将列表转换为向量。
#include <list>
#include <vector>
list<Object> myList;
vector<Object> myVector(myList.begin(), myList.end());
Then access the vector using the [x] operator.
然后使用 [x] 运算符访问向量。
auto x = MyVector[0];
You could put that in a helper function:
你可以把它放在一个辅助函数中：
#include <memory>
#include <vector>
#include <list>
template<class T>
shared_ptr<vector<T>>
ListToVector(list<T> List) {
shared_ptr<vector<T>> Vector {
new vector<string>(List.begin(), List.end()) }
return Vector;
}
Then use the helper funciton like this:
然后像这样使用辅助函数：
auto MyVector = ListToVector(Object);
auto x = MyVector[0];
