How to increment an iterator by 2?

问题:

Can anybody tell me how to increment the iterator by 2?

iter++ is available - do I have to do iter+2? How can I achieve this?

回答1:

std::advance( iter, 2 );

This method will work for iterators that are not random-access iterators but it can still be specialized by the implementation to be no less efficient than iter += 2 when used with random-access iterators.

回答2:

http://www.cplusplus.com/reference/std/iterator/advance/

std::advance(it,n);

where n is 2 in your case.

The beauty of this function is, that If "it" is an random access iterator, the fast

it += n

operation is used (i.e. vector<,,>::iterator). Otherwise its rendered to

for(int i = 0; i < n; i++)
    ++it;

(i.e. list<..>::iterator)

回答3:

If you don't have a modifiable lvalue of an iterator, or it is desired to get a copy of a given iterator (leaving the original one unchanged), then C++11 comes with new helper functions - std::next / std::prev:

std::next(iter, 2);          // returns a copy of iter incremented by 2
std::next(std::begin(v), 2); // returns a copy of begin(v) incremented by 2
std::prev(iter, 2);          // returns a copy of iter decremented by 2

回答4:

You could use the 'assignment by addition' operator

iter += 2;

回答5:

If you don't know wether you have enough next elements in your container or not, you need to check against the end of your container between each increment. Neither ++ nor std::advance will do it for you.

if( ++iter == collection.end())
  ... // stop

if( ++iter == collection.end())
  ... // stop

You may even roll your own bound-secure advance function.

If you are sure that you will not go past the end, then std::advance( iter, 2 ) is the best solution.

回答6:

We can use both advance as well as next. But, there's a difference between the two. "advance" modifies its argument and returns nothing.So, it can be used as:

vector<int> v;
v.push_back(1);
v.push_back(2);
auto itr = v.begin();
advance(itr, 1);           //modifies the itr
cout << *itr<<endl        //prints 2

"next" returns a modified copy of the iterator

vector<int> v;
v.push_back(1);
v.push_back(2);
cout << *next(v.begin(), 1) << endl;    //prints 2

回答7:

Assuming list size may not be an even multiple of step you must guard against overflow:

static constexpr auto step = 2;

// Guard against invalid initial iterator.
if (!list.empty())
{
    for (auto it = list.begin(); /*nothing here*/; std::advance(it, step))
    {
        // do stuff...

        // Guard against advance past end of iterator.
        if (std::distance(it, list.end()) > step)
            break;
    }
}

Depending on the collection implementation, the distance computation may be very slow. Below is optimal and more readable. The closure could be changed to a utility template with the list end value passed by const reference:

const auto advance = [&](list_type::iterator& it, size_t step)
{
    for (size_t i = 0; it != list.end() && i < step; std::next(it), ++i);
};

static constexpr auto step = 2;

for (auto it = list.begin(); it != list.end(); advance(it, step))
{
    // do stuff...
}

If there is no looping:

static constexpr auto step = 2;
auto it = list.begin();

if (step <= list.size())
{
    std::advance(it, step);
}

回答8:

The very simple answer:

++++iter

The long answer:

You really should get used to writing ++iter instead of iter++. The latter must return (a copy of) the old value, which is different from the new value; this takes time and space.

Note that prefix increment (++iter) takes an lvalue and returns an lvalue, whereas postfix increment (iter++) takes an lvalue and returns an rvalue.