如何获取列表中的某个元素,给定位置? [英] How to get a certain element in a list, given the position?
问题描述
所以我有一个列表:
list<Object> myList;
myList.push_back(Object myObject);
我不确定,但我相信这将是数组。
是否有任何函数可以使用将返回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"?
Object copy = myList.find_element(0);
?
推荐答案
如果你经常需要访问序列的第N个元素,那么作为双向链表实现的 std :: list
可能不是正确的选择。 std :: vector
或 std :: deque
可能会更好。
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::vector
or std::deque
would likely be better.
也就是说,你可以使用 std :: advance
获得迭代器到第N个元素:
That said, you can get an iterator to the Nth element using 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'
}
对于不提供随机访问的容器,如 std :: list
, std :: advance
运算符++
在迭代器 N
次。或者,如果您的标准库实现提供它,您可以调用 std :: next
:
For a container that doesn't provide random access, like std::list
, std::advance
calls operator++
on the iterator N
times. Alternatively, if your Standard Library implementation provides it, you may call std::next
:
if (l.size() > N)
{
std::list<Object>::iterator it = std::next(l.begin(), N);
}
std :: next
有效地调用 std :: advance
,使得更容易推进迭代器 N
次的代码和较少的可变变量。 std :: next
已在C ++ 11中添加。
std::next
is effectively wraps a call to std::advance
, making it easier to advance an iterator N
times with fewer lines of code and fewer mutable variables. std::next
was added in C++11.
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