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std::erase, std::erase_if(std::inplace_vector) - cppreference.com

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Defined in header <inplace_vector> 

template< class T, std::size_t N, class U = T >
constexpr typename std::inplace_vector<T, N>::size_type
    erase( std::inplace_vector<T, N>& c, const U& value );
(1) (since C++26) 
template< class T, std::size_t N, class Pred >
constexpr typename std::inplace_vector<T, N>::size_type
    erase_if( std::inplace_vector<T, N>& c, Pred pred );
(2) (since C++26)

1) Erases all elements that compare equal to value from the container. Equivalent to 

auto it = std::remove(c.begin(), c.end(), value);
auto r = std::distance(it, c.end());
c.erase(it, c.end());
return r;

2) Erases all elements that satisfy the predicate pred from the container. Equivalent to 

auto it = std::remove_if(c.begin(), c.end(), pred);
auto r = std::distance(it, c.end());
c.erase(it, c.end());
return r;

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 v of type (possibly const) T, regardless of value category, and must not modify v. Thus, a parameter type of T&is not allowed, nor is T unless for T a move is equivalent to a copy(since C++11). ​

Return value

The number of erased elements.

Complexity

Linear.

Example

#include <algorithm>
#include <cassert>
#include <complex>
#include <inplace_vector>
#include <numeric>
#include <print>
#include <vector>

int main()
{
    std::inplace_vector<int, 10> v(10, 0);
    std::ranges::iota(v, 0);
    std::println("Initially, v = {}", v);
    
    auto erased = std::erase(v, 3);
    std::println("After erase(v, 3), v = {}", v);
    assert(erased == 1);
    
    erased = std::erase_if(v, [](int x) { return x % 2 == 0; });
    std::println("After erasing all even numbers, v = {}", v);
    std::println("Erased even numbers: {}", erased);
    
    std::inplace_vector<std::complex<double>, 4> nums{{2, 2}, {4, 2}, {4, 8}, {4, 2}};
    std::erase(nums, {4, 2}); // since U = T, the value type can be omitted
    assert(std::ranges::equal(nums, std::vector<std::complex<double>>({{2, 2}, {4, 8}})));
}

Output: 

Initially, v = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
After erase(v, 3), v = [0, 1, 2, 4, 5, 6, 7, 8, 9]
After erasing all even numbers, v = [1, 5, 7, 9]
Erased even numbers: 5

See also