std::list<T,Allocator>::erase

Since C++11

Erases the specified elements from the container.

# Declarations

iterator erase( iterator pos );

(until C++11)

iterator erase( const_iterator pos );

(since C++11)

iterator erase( iterator first, iterator last );

(until C++11)

iterator erase( const_iterator first, const_iterator last );

(since C++11)

# Parameters

pos: iterator to the element to remove
first, last: range of elements to remove

# Return value

Iterator following the last removed element.

# Notes

When container elements need to be erased based on a predicate, rather than iterating the container and calling unary erase, the iterator range overload is generally used with std::remove()/std::remove_if() to minimise the number of moves of the remaining (non-removed) elements, — this is the erase-remove idiom. std::erase_if() replaces the erase-remove idiom.(since C++20)

# Example

#include <list>
#include <iostream>
#include <iterator>
 
void print_container(const std::list<int>& c)
{
    for (int i : c)
        std::cout << i << ' ';
    std::cout << '\n';
}
 
int main()
{
    std::list<int> c{0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
    print_container(c);
 
    c.erase(c.begin());
    print_container(c);
 
    std::list<int>::iterator range_begin = c.begin();
    std::list<int>::iterator range_end = c.begin();
    std::advance(range_begin, 2);
    std::advance(range_end, 5);
 
    c.erase(range_begin, range_end);
    print_container(c);
 
    // Erase all even numbers
    for (std::list<int>::iterator it = c.begin(); it != c.end();)
    {
        if (*it % 2 == 0)
            it = c.erase(it);
        else
            ++it;
    }
    print_container(c);
}

# Defect reports

DR	Applied to	Behavior as published	Correct behavior
LWG 151	C++98	first was required to be dereferenceable, whichmade the behavior of clearing an empty list undefined	not required iffirst == last