## Are Iterators to Erased Elements in a `std::vector` Still Valid?

std::vector Iterator Invalidation: Unearthing the Validity of Iterators After Erasure

In the realm of C , the intricacies of vector iterators and their behavior after erasures can be a source of confusion. To delve into this issue, we'll explore a specific scenario:

Validating Iterator Persistence After an Erase Operation

Pertinent questions arise regarding the validity of an iterator pointing directly to the erased element in a std::vector. The prevailing notion is that iterators pointing to positions after the erased element are invalidated. However, the question remains: is the iterator pointing to the exact location of the erased element still valid?

Analysis and Example

To shed light on this issue, consider the following code snippet, which attempts to eliminate all odd integers from a vector:

<code class="cpp">vector<int> vec;

for (int i = 0; i < 100; ++i) vec.push_back(i);

vector<int>::iterator it = vec.begin();
while (it != vec.end()) {
    if (*it % 2 == 1) vec.erase(it);
    else ++it;
}</code>

While this code may appear to function correctly, it's crucial to delve deeper to determine its validity.

Answering the Question

The answer unequivocally points to the invalidation of not only iterators pointing to positions after the erased element but also the iterator pointing to the exact location of that element.

However, erasing an element provides a returned iterator pointing immediately after the removed element(s) or to the end if there are none remaining. This allows for seamless continuation of iteration.

Efficient Elimination of Odd Elements

It's worth noting that the code snippet presented is not the most efficient method for removing odd elements. A significantly more efficient approach involves the erase-remove idiom, which utilizes a custom predicate to locate and identify elements for removal. For example, we can define a predicate called is_odd and use it with remove_if and erase:

<code class="cpp">bool is_odd(int x) { return (x % 2) == 1; }
vec.erase(remove_if(vec.begin(), vec.end(), is_odd), vec.end());</code>

This eliminates the costly movement of elements, reducing the time complexity to O(n).

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