std::list::remove, std::list::remove_if

 

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void remove( const T& value );
template< class UnaryPredicate > void remove_if( UnaryPredicate p );

Removes all elements satisfying specific criteria. The first version removes all elements that are equal to value, the second version removes all elements for which predicate p returns true.

Parameters

value	-	value of the elements to remove
p	-	unary predicate which returns ​true if the element should be removed. The signature of the predicate function should be equivalent to the following: bool pred(const Type &a); The signature does not need to have const &, but the function must not modify the objects passed to it. The type Type must be such that an object of type list<T,Allocator>::const_iterator can be dereferenced and then implicitly converted to Type. ​

Return value

(none)

Complexity

linear in the size of the container

Example

run this code

#include <list>
#include <iostream>
 
int main()
{
    std::list<int> l = { 1,100,2,3,10,1,11,-1,12 };
 
    l.remove(1); // remove both elements equal to 1
    l.remove_if([](int n){ return n > 10; }); // remove all elements greater than 10
 
    for (int n : l) {
        std::cout << n << ' '; 
    }
    std::cout << '\n';
}

Output:

2 3 10 -1

See also

remove remove_if	removes elements satisfying specific criteria (function template)