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Programming-Idioms

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  • Java

Idiom #247 Filter list in-place

Remove all the elements from list x that don't satisfy the predicate p, without allocating a new list.
Keep all the elements that do satisfy p.

For languages that don't have mutable lists, refer to idiom #57 instead.

x.removeIf(p.negate());
#include <list>
#include <functional>
std::list<Foo> x;
x.remove_if(std::not_fn(p));

C++17 or later
Also works if x is an std::forward_list

New implementation...