std::{{{1}}}::erase

 

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	(1)
iterator erase( iterator pos );			(until C++11)
iterator erase( const_iterator pos );			(since C++11)
		(2)
iterator erase( iterator first, iterator last );				(until C++11)
iterator erase( const_iterator first, const_iterator last );				(since C++11)

Erases the specified elements from the container.

1) Removes the element at pos.

2) Removes the elements in the range [first, last).

The iterator pos must be valid and dereferenceable. Thus the end() iterator (which is valid, but is not dereferencable) cannot be used as a value for pos.

The iterator first does not need to be dereferenceable if first==last: erasing an empty range is a no-op.

Parameters

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

Return value

Iterator following the last removed element. If the iterator pos refers to the last element, the end() iterator is returned.

Example

Run this code

#include <{{{1}}}>
#include <iostream>
#include <iterator>
 
int main( )
{
    std::{{{1}}}<int> c{0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
    for (auto &i : c) {
        std::cout << i << " ";
    }
    std::cout << '\n';
 
    c.erase(c.begin());
 
    for (auto &i : c) {
        std::cout << i << " ";
    }
    std::cout << '\n';
 
    std::{{{1}}}<int>::iterator range_begin = c.begin();
    std::{{{1}}}<int>::iterator range_end = c.begin();
    std::advance(range_begin,2);
    std::advance(range_end,5);
 
    c.erase(range_begin, range_end);
 
    for (auto &i : c) {
        std::cout << i << " ";
    }
    std::cout << '\n';
 
    // Erase all even numbers (C++11 and later)
    for (auto it = c.begin(); it != c.end(); ) {
        if (*it % 2 == 0) {
            it = c.erase(it);
        } else {
            ++it;
        }
    }
 
    for (auto &i : c) {
        std::cout << i << " ";
    }
    std::cout << '\n';
}

Output:

0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 6 7 8 9
1 7 9

See also

clear

clears the contents (public member function of std::{{{1}}})