STL list, delete all the odd numbers

Question

I am trying to learn how to work with STL and tried to write a function which will recieve a refference to a list and will try to delete all odd members. I am having a sligh

Answer 1

Method erase returns iterator to the next element after deleted. So the usual approach to such a task is the following

void removeOdds( std::list<int> &myvector )
{
    for ( auto it = myvector.begin(); it != myvector.end();  )
    {
        if ( *it % 2 !=0 )
        {
            it = myvector.erase( it );
        }
        else
        {
            ++it;
        }
    }
}

As for your code then the loop statement always increases the iterator

for(list<int>::iterator p=myvector.begin(); p !=myvector.end();p++)
                                                               ^^^^  

Take into account that class std::list has methods remove and remove_if that erase all elements that satisfy the given criteria.

Here is a demonstrative program that shows the both approaches

#include <iostream>
#include <list>

void removeOdds( std::list<int> &myvector )
{
    for ( auto it = myvector.begin(); it != myvector.end();  )
    {
        if ( *it % 2 !=0 )
        {
            it = myvector.erase( it );
        }
        else
        {
            ++it;
        }
    }
}

int main()
{
    std::list<int> l = { 1, 3, 5, 2, 6, 7 };

    for ( auto x : l ) std::cout << x << ' ';
    std::cout << std::endl;

    removeOdds( l );

    for ( auto x : l ) std::cout << x << ' ';
    std::cout << std::endl;

    l = { 1, 3, 5, 2, 6, 7 };

    for ( auto x : l ) std::cout << x << ' ';
    std::cout << std::endl;

    l.remove_if( []( int x ) { return x % 2; } );

    for ( auto x : l ) std::cout << x << ' ';
    std::cout << std::endl;
}

The program output is

1 3 5 2 6 7 
2 6 
1 3 5 2 6 7 
2 6 

Answer 2

Since you're learning, you better forget the 'C' way of doing things and jump directly into C++11.

Here's your code in modern C++

void removeOdds(vector<int>& myvector)
{
    auto new_end = std::remove_if (myvector.begin(), myvector.end(), 
     [](const auto& value){ return value%2 !=0 }
    );    
    myvector.erase(new_end, myvector.end());
}

As far as your code goes, you can improve readability a lot just by using auto Please not that it is risky to remove elements while looping on a container. You're usually fine with a vector, but when you start using other data structures things get more nasty.

Answer 3

Ok, I found one way to do this, and that was to move p++ from for loop declaration into inside it, like this.

void removeOdds(list<int>& myvector)
{
    for(list<int>::iterator p=myvector.begin(); p !=myvector.end();)
    {
        if(*p%2 !=0)
        {
            list<int>::iterator temp=myvector.erase(p);
            p=temp;
        }
        else
            p++;
    }
} 

Now I am curious if I can do this and still keep my for loop intact.

Answer 4

The easiest way to do it would be to use std::list::remove_if. This removes elements from the list based on a unary predicate. For example,

myvector.remove_if([](int n) { return n % 2 != 0; });

The best way to work with the "STL"^* is to know what is in it.

For pre-C++11 implementations (such as the actual STL), you can pass a function:

bool is_odd(int n) { return n % 2 != 0; }

myvector.remove_if(is_odd);

---

^{*"STL" means the STL but this also applies the C++ standard library}