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Usually in C++ one can iterate over each second element by doing so:

for (int i = 0; i < 10; i = i + 2 ){ 
       //do Somthing 
}

and for a C++ list (from library list) one can iterate over each element like so:

std::list<int> dY = {5,4,5,8,9,7,10,4};
std::list<int>::iterator iterdY;
for (iterdY = dY.begin(); iterdY != dY.end(); iterdY++) {
        // do Something
    }

Now my question is what should I write instead of iterdY++ to iterate over each second element ??

M.Zayed
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  • Could you not just use a regular `for` loop, without the iterators? (Defeats the point of the question, but might be helpful) – cs1349459 Oct 03 '22 at 15:56
  • @cs1349459 No raw for loops! https://youtu.be/f_tLQl0wLUM?t=583 – Pepijn Kramer Oct 03 '22 at 15:57
  • Does this answer your question? [How to increment an iterator by 2?](https://stackoverflow.com/questions/1057529/how-to-increment-an-iterator-by-2) Just be careful about containers that have an odd number of elements. It's up to you to prevent incrementing your iterator past the end. – Wyck Oct 03 '22 at 16:20

5 Answers5

3

Probably the easiest way is to introduce a second iteration variable.

std::list<int> dY;
int n = 0;
std::list<int>::iterator iterdY = {5,4,5,8,9,7,10,4};
for (iterdY = dY.begin(); iterdY != dY.end(); iterdY++, ++n) {
    if ((n % 2) == 0) {
        // do Something
    }
}
Sven Nilsson
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2

std::advance(iterdY, 2)

But you can no longer use the comparison iterdY != dY.end() because if the list has an odd number of elements you will skip over the end iterator, and into the territory of undefined behavior.

You must make sure that the list have an even number of elements first, if using iterators, or use some other way to keep track of the iterator not passing beyond the end.

If your compiler supports C++23 you can use ranges and stride_view to skip over every second element in the container.

See example here (using the earlier Ranges v3 library, which is the base for the C++20 standard library ranges).

Using the original ranges v3 library with earlier C++ standards is possible as well.

Some programmer dude
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2

For starters I think you mean a list in this declaration

std::list<int>::iterator iterdY = {5,4,5,8,9,7,10,4};

like

std::list<int> dY = {5,4,5,8,9,7,10,4};

instead of the iterator.

You can do something like the following

for ( auto it = dY.begin(); it != dY.end(); ++it == dY.end() ? it : ++it ) {
//...

Here is a demonstration program.

#include <iostream>
#include <list>

int main() 
{
    std::list<int>dY = {5,4,5,8,9,7,10,4}; 

    for ( auto it = dY.begin(); it != dY.end(); ++it == dY.end() ? it : ++it )
    {
        std::cout << *it << ' ';
    }
    std::cout << '\n';       
}

The program output is

5 5 9 10
Vlad from Moscow
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I'd use two steps and a function:

auto iterdY = dY.begin();
while (iterDy != dy.end()) {
    do_something(*iterDy++);
    if (iterDy == dY.end())
        break;
    do_something(*iterDy++);
}
Pete Becker
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0

After understanding all of suggested answers I would like to propose this solution:

#include <iostream>
#include <string>
#include <list>
using namespace std;

int main()
{
    std::list<int> dY = {5,4,5,8,9,10,11}; 
    std::list<int>::iterator it = dY.begin();
    int step = 3; // Here one can change the step to print each third element
    for (int i = 0 ; i < dY.size() ; i+= step )
    {
        std::cout << *it <<'\n';
        std::advance(it, step);
    } 
    std::cout << '\n';       
}

output is:

5 8 11

M.Zayed
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