std::erase, std::erase_if (std::list)
From cppreference.com
Defined in header <list>
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(1) | (since C++20) | |
(2) | (since C++20) | |
1) Erases all elements that compare equal to
value
from the container. Equivalent to return c.remove_if([&](auto& elem) { return elem == value; });2) Erases all elements that satisfy the predicate
pred
from the container. Equivalent to return c.remove_if(pred);Parameters
c | - | container from which to erase |
value | - | value to be removed |
pred | - | unary predicate which returns true if the element should be erased. The expression pred(v) must be convertible to bool for every argument |
Complexity
Linear.
Example
Run this code
#include <iostream> #include <numeric> #include <list> void print_container(const std::list<char>& c) { for (auto x : c) { std::cout << x << ' '; } std::cout << '\n'; } int main() { std::list<char> cnt(10); std::iota(cnt.begin(), cnt.end(), '0'); std::cout << "Init:\n"; print_container(cnt); std::erase(cnt, '5'); std::cout << "Erase \'5\':\n"; print_container(cnt); std::erase_if(cnt, [](char x) { return (x - '0') % 2 == 0; }); std::cout << "Erase all even numbers:\n"; print_container(cnt); }
Output:
Init: 0 1 2 3 4 5 6 7 8 9 Erase '5': 0 1 2 3 4 6 7 8 9 Erase all even numbers: 1 3 7 9
Notes
Unlike std::list::remove, erase
accepts heterogenous types and does not force a conversion to the container's value type before invoking the == operator.
See also
removes elements satisfying specific criteria (function template) | |
removes elements satisfying specific criteria (public member function) |