loop through a vector using iterator and test for end of loop-

Natural_Bugger · M mrjj 26 Jul 2020, 09:28

@mrjj
hi and thnx,
but how do i know i reached the last "iteration" of the loop?
the last element of "it".

if(......){
    // last element of "it"
   }

Chris Kawa · 26 Jul 2020, 09:48

if (it == std::prev(mylist.end()))

Christian Ehrlicher · 26 Jul 2020, 10:03

@Natural_Bugger said in loop through a vector using iterator and test for end of loop-:

but how do i know i reached the last "iteration" of the loop?

You can do it exactly the same as in a for loop which a loop variable

if (it == myList.end() - 1)

or use the solution from @Chris-Kawa

Natural_Bugger · wrote on 26 Jul 2020, 09:48

@Chris-Kawa

thnx

Natural_Bugger · wrote on 26 Jul 2020, 10:03

@Christian-Ehrlicher

thnx

Natural_Bugger · wrote on 26 Jul 2020, 10:30

@Chris-Kawa
@Christian-Ehrlicher

i wanna remove the current element i'm "testing"

http://www.cplusplus.com/reference/vector/vector/erase/

iterator erase (iterator position);
iterator erase (iterator first, iterator last);

std::cout << "*it: " << *it << std::endl; // outputs the correct  numbers
mylist.erase(*it);

error:

error: no matching function for call to ‘std::vector<int>::erase(int&)’

i also tried:

mylist..erase(ar.begin() + *it);

results is fatal crash.

Chris Kawa · 26 Jul 2020, 11:15

*it is the value pointed to by the iterator. ar.begin() + *it is nonsense. It advances the iterator by value pointed to by it.

You want to remove the element at iterator so simply

mylist.erase(it);

Keep in mind that erasing the element invalidates the iterator, so if you're doing it in a loop you need to use the iterator returned by erase:

it = mylist.erase(it);

which erases the element pointed by it and then sets it to point to next element.

Natural_Bugger · wrote on 26 Jul 2020, 11:15

@Chris-Kawa
thnx,
ah ... removing the "*", i dind't try that.
at first when i tried to use "if else" statements inside the loop, the compiler couldn't convert "int" into pointer.
so, i though i had to use it.

if(mylistr[0] == *it){
                    std::cout << "*it: " << *it << std::endl; // i do need to use pointer here
                    mylist.erase(mylist.begin()); // remove first
                    mylist.erase(it); // remove current.
                break;
                }

i'm existing the loop after i remove the element.

break;

but i still have error on certain conditions, using a list with filled different values, some are valid, some crash.

Program terminated with signal SIGSEGV, Segmentation fault.

Natural_Bugger · wrote on 26 Jul 2020, 11:29

@Chris-Kawa

this is how my loops looks like ...

find a pair of matching numbers and remove them.
the find next pair ...
etc

int pairs = 0;
int counter = 0;

do{
        if(counter == n){ // n = number of number in list.
            break;
        }
int counterB = 0;
bool remove = false;
for(std::vector<int>::iterator it = (mylist.begin()+ 1); it != mylist.end(); ++it) {
                if(mylist[0] == *it){
                    std::cout << "*it: " << *it << std::endl;
                    mylist.erase(mylist.begin());
                    mylist.erase(it); // fails occasionally, but more the the line below.
                    // mylist..erase(mylist.begin() + counterB); // works 100% ... but does not pas all tests.
                    pairs++;
                break;
                }
                if(it == std::prev(mylist.end())){
                    remove = true; // bool value
                }
                   counterB++;
        }
if(remove){
            mylist.erase(mylist.begin()); // value didn't occur more than once, remove first value.
            remove = false;
        }
while(true);
...
..
.

kind regards.

Chris Kawa · 26 Jul 2020, 11:45

@Natural_Bugger said:

mylist.erase(mylist.begin());
mylist.erase(it); // fails occasionally, but more the the line below.

As I said in my previous post erase invalidates iterators so it in the second line points to garbage. For these two lines it's enough to switch them, but you're using it couple lines lower and that's also an error. I repeat - whenever you use erase any iterator you have is invalid. erase returns an iterator to element after the one that was removed. Keep in mind that the removed element could be the last one so iterator returned from erase can be end() so pointing past the container.

By the way, if all you want to do is remove duplicates you can just do:

std::sort(mylist.begin(), mylist.end());
mylist.erase(std::unique(mylist.begin(), mylist.end()), mylist.end() );

Natural_Bugger · wrote on 26 Jul 2020, 11:45

@Chris-Kawa

how do i store the position of the element, found by the condition of the statement and remove it outside the loop.

if(mylistr[0] == *it){
    ...
   ..
   .
                break;
  }

regards.

Natural_Bugger · wrote on 26 Jul 2020, 11:56

@Chris-Kawa

thnx

std::sort(mylist.begin(), mylist.end());
mylist.erase(std::unique(mylist.begin(), mylist.end()), mylist.end() );

i need to find a pair, count it, remove that pair, find next pair, count it, etc, etc ... but not all number in the list have a pair.

well ... i just need to count the pairs ... somehow,
so i think the best way to do that, is to find a pair and remove it, so you don't keep finding it over and over.

Natural_Bugger · wrote on 26 Jul 2020, 13:37

@Chris-Kawa

in the end, i changed the code to store the positions and remove it outside the loop.
worked fine, but the online compiler didn't pas all tests.
than i tried with the Qt creator IDE and the code return good results.

now, I'm trying that what you suggesting.

std::sort(mylist.begin(), mylist.end());

but how to turn

vector::begin
vector::end

into a INT values to compare against?
so i don't over run the max length of the list.

the code became much much smaller, but also tries to compare value beyond the end of the vector.

   int pairs = 0;
    sort( mylist.begin(), mylist.end() );
    int counter = 0;
    for(auto ii : mylist){
        if(mylist.at(counter) == mylist.at(counter + 1)){
            std::cout << "found duplicate" << std::endl;
            pairs++;
        } 
        counter ++;
    }

Natural_Bugger · 27 Jul 2020, 08:49

intput:

20 // number of value
4 5 5 5 6 6 4 1 4 4 3 6 6 3 6 1 4 5 5 5

sorting:

std::sort(mylist.begin(), mylist.end());

atfter sorting:

1 1 3 3 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6

int example(vector<int> mylist){
sort( ar.begin(), ar.end() );
int pairs = 0;
unsigned long int counterC = 0;
    do{

        if(counterC == (mylist.size() - 1)){
            std::cout << "end loop" << std::endl;
            break;
        }
        else
        {
            do
            {
                if(mylist[0] == mylist[1])
                {
                    // equal;
                    mylist.erase(mylist.begin()); // erase first element, than try to erase next ...... it works ... but partially ... at first iteration it does remove the first 2 element, but not on the following.
                    mylist.erase(mylist.begin()); // breaks "sometimes" the code according to Qt creator debug
                    // Program terminated with signal SIGSEGV, Segmentation fault.
                    std::cout << "found equals" << std::endl;
                    pairs++;
                    break;
                }
                else
                {
                    // remove
                    std::cout << "found soletary" << mylist.at(0) << std::endl;
                    ar.erase(ar.begin());
                    break;
                }
            }
            while(true);
        }
        for(auto ii : mylist)
        {
            std::cout << "ii: "<< ii << " ";
        }
        std::cout << std::endl;
        std::cout << "counterC: " << counterC << std::endl;
        counterC++;
    }
    while(true);
return pairs;
}

i also tried:

mylist.erase(mylist.begin(), mylist.begin()+1);

this is the total output:

found equals
ii: 1 ii: 3 ii: 3 ii: 4 ii: 4 ii: 4 ii: 4 ii: 4 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 0
found soletary1
ii: 3 ii: 3 ii: 4 ii: 4 ii: 4 ii: 4 ii: 4 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 1
found equals
ii: 3 ii: 4 ii: 4 ii: 4 ii: 4 ii: 4 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 2
found soletary3
ii: 4 ii: 4 ii: 4 ii: 4 ii: 4 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 3
found equals
ii: 4 ii: 4 ii: 4 ii: 4 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 4
found equals
ii: 4 ii: 4 ii: 4 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 5
found equals
ii: 4 ii: 4 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 6
found equals
ii: 4 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 7
found soletary4
ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 8
found equals
ii: 5 ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 9
found equals
ii: 5 ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 10
found equals
ii: 5 ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 11
found equals
ii: 5 ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 12
found equals
ii: 5 ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 13
found soletary5
ii: 6 ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 14
found equals
ii: 6 ii: 6 ii: 6 ii: 6 
counterC: 15
found equals
ii: 6 ii: 6 ii: 6 
counterC: 16
found equals
ii: 6 ii: 6 
counterC: 17
found equals
ii: 6 
counterC: 18
found equals

counterC: 19

JonB · 27 Jul 2020, 11:26

@Natural_Bugger
In some shape or form: when iterating through the elements looking at adjacent ones, you must either omit/skip the first and start at the second one (if you compare elements against the one to the left), or omit/skip the last one and stop at the penultimate one (if you compare elements against the one to the right).

Natural_Bugger · J JonB 27 Jul 2020, 08:49

@JonB

thank you for the reply.

well, i saw that some people or most did that for the same problem, but in Java and apparently, they also sorted the "array/map" like Chris Kawa mentioned it worked for them.
so i tried that as well in c++.

bit i also think the "solutions" they published do not work, because they only compare 2 (if) or else (nothing). no taking in consideration if the list of values is either even or uneven in length or the numbers of equal value is either even or uneven.

now i'm using nested "Do While" loops instead of For loop in combination with Iteration.
what to do next?

.