Home » C++ » How to get a certain element in a list, given the position?

How to get a certain element in a list, given the position?

Posted by: admin November 30, 2017 Leave a comment

Questions:

So I’ve got a list:

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”?

Object copy = myList.find_element(0);

?

Answers:

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.

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'
}

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 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.

Questions:
Answers:

std::list doesn’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.

What you need is : std::vector. Use it as:

std::vector<Object> objects;
objects.push_back(myObject);

Object obj = objects[0]; //get element given an index

Questions:
Answers:
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
}