std::remove but specify max elements to remove?

[rozick1]

std::remove will 'remove' every matching element in a container, how can I specify that it removes up to a max number of matching elements?

[dutch]

There's no way to get std::remove to do that, but you can do it using std::remove_if:

#include <iostream>
#include <vector>
#include <algorithm>

int main()
{
    const int MaxRemove = 3;

    std::vector<int> v{ 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1 };
    for ( int n: v ) std::cout << n << ' '; std::cout << '\n';
    std::cout << std::count( v.begin(), v.end(), 1 ) << " 1's\n";

    int n = 0;
    v.erase( std::remove_if( v.begin(), v.end(),
                             [&](auto x){ return x == 1 && n++ < MaxRemove; } ),
             v.end() );

    for ( int n: v ) std::cout << n << ' '; std::cout << '\n';
    std::cout << std::count( v.begin(), v.end(), 1 ) << " 1's\n";
}

[jonnin]

can you do a remove-if with some sort of counter?
you may also be able to tamper with the iterator from remove so that when you call erase it only picks up to N?

[lastchance]

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

template<typename T>
void removeSome( vector<T> &V, T value, int n, int i = 0 )
{
   if ( n <= 0 || i >= V.size() ) return;
   auto it = find( V.begin() + i, V.end(), value );
   if ( it == V.end() ) return;
   i = it - V.begin();
   rotate( it, it + 1, V.end() );
   V.pop_back();
   removeSome( V, value, n - 1, i );
}


template<typename T>
ostream & operator << ( ostream &out, const vector<T> &V )
{
   for ( T e : V ) out << e << ' ';
   return out;
}


int main()
{
    const int MaxRemove = 3, Value = 1;
    vector<int> V{ 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1 };

    cout << "Before: " << V << '\n';
    removeSome( V, Value, MaxRemove );
    cout << "After: " << V << '\n';
}