std::deque erase() method
- od C++11
- do C++11
// (1) Non const version only
iterator erase( const_iterator pos );
// (2) Non const version only
iterator erase( const_iterator first, const_iterator last );
// (1) Non const version only
iterator erase( iterator pos );
// (2) Non const version only
iterator erase( iterator first, iterator last );
Erases the specified elements from the container.
- (1) Removes the element at
pos
. - (2) Removes the elements in the range [ first, last ).
All iterators and references are invalidated, unless the erased elements are at the end or the beginning of the container, in which case only the iterators and references to the erased elements are invalidated.
- od C++11
- do C++11
The iterator pos
must be valid and dereferenceable.
Thus the end()
iterator (which is valid, but is not dereferenceable) cannot be used as a value for pos
.
For overload (2), 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 removefirst
,last
- range of elements to remove
Type requirements
T
(the container's element type) must meet the requirements ofMoveAssignable
.
Return value
Iterator following the last removed element.
- For (1), if
pos
refers to the last element, then theend()
iterator is returned. - For (2), if
last == end()
prior to removal, then the updatedend()
iterator is returned. - For (2), if [ first, last ) is an empty range, then
last
is returned.
Complexity
Linear: the number of calls to the destructor of T
is the same as the number of elements erased,
the number of calls to the assignment operator of T
is no more than the lesser of the number of elements before the erased elements and the number of elements after the erased elements.
Exceptions
Does not throw, unless an exception is thrown by the assignment operator of T
.
Example
#include <deque>
#include <iostream>
void print_container(const std::deque<int>& c)
{
for (int i : c)
std::cout << i << " ";
std::cout << '\n';
}
int main( )
{
std::deque<int> c{0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
print_container(c);
c.erase(c.begin());
print_container(c);
c.erase(c.begin()+2, c.begin()+5);
print_container(c);
// Erase all even numbers
for (std::deque<int>::iterator it = c.begin(); it != c.end();)
{
if (*it % 2 == 0)
it = c.erase(it);
else
++it;
}
print_container(c);
}
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
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