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runtime Error Erase element vector C++

Posted on 2013-06-14
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Last Modified: 2013-06-14
Hello,
I have a
vector<int>Index = [1,3,4,5,8,9]

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vector<int>numbers = [1,2,3,4,5,6,7,8,9,10]

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I want to use the indices in the vector index to remove the numbers in vector<int>numbers
Example:
for(int t1 = 0; t1 < index.size(); t1++)
{
        words.erase(numbers.begin()+ index[t1])
}

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But my program keeps on crashing.
I tried the reverse way also :
for(int t1 = index.size()-1; t1 > 0; --t1)
{
    words.erase(words.begin()+index[t1])
}

I also tried to decrement the values in the vector Index since every time i remove an element from vector numbers the index will decrease right?
So also it crashed
 int counter = 0;
for(int t1 = index.size()-1; t1 > 0; t1--)
   {
      //  cout << index[t1]<<endl;
       words.erase(words.begin() +( index[t1]-counter));
       counter = counter + 1;
   }

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0
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Question by:HaniDaher
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4 Comments
 
LVL 37

Expert Comment

by:TommySzalapski
ID: 39247275
You had the right idea, you just solved it both ways.
Either going through the array backwards OR subtracting the counter would have worked, but when you do both, you run off the front of the array.

Going through the array backwards is the most standard way to do it. So just remove the 'counter' variable and it should work.
0
 

Author Comment

by:HaniDaher
ID: 39247304
Hello Tommy,  Thank you for your answers as usual.
Tommy I tried this but it didn't work i'll tell you why.

Suppose that i have the vector<int> INDEX = {5,6,7}
and I have the vector<int> Letters = {A,B,C,D,E,F,G,H,I,J,K,L}
I want to remove the letters at index 5, 6 and 7 right?
So  When i remove the letter at Index 5 which is "F"
Letter = {A,B,C,D,E,G,H,I,J,K,L} And in vector INDEX: 6 should become 5 and 7 should become 6.

What do you think? I'll post this code and let me know if it's right or if my logic is wrong.

    vector<int>a = {0,0,0,0,1,1,1,1,0,0,0,0,1,0,0,0,0,0,0};
    vector<int>index;
    int t = 0;
    while(t < a.size() && a[t] != 1)
    {
        index.push_back(t);
        t = t + 1;
    }

    t = a.size()-1;
    while( t > 0 && a[t] != 1)
    {
        index.push_back(t);
        t = t - 1;
    }
    sort(index.begin(),index.end());
    index.erase(std::unique(index.begin(), index.end()), index.end());
    cout << "Before: ";
    for(int i = 0; i <a.size();i++)
    {
       cout << a[i] <<"   ";
    }


    cout << endl;
    int counter = 0;
    for(int i = 0; i < index.size();i++)
    {
        a.erase(a.begin() + (index[i]-counter));
        counter = counter + 1;
    }

    cout <<"After: ";
    for(int i = 0; i < a.size();i++)
    {
       cout << a[i] <<"   ";
    }

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0
 
LVL 37

Accepted Solution

by:
TommySzalapski earned 2000 total points
ID: 39247372
Yes, but if you do it like this:
for(int t1 = index.size()-1; t1 >= 0; t1--)
   {
       words.erase(words.begin() +( index[t1]));
   }

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Then it should work. Because when you erase the element 7, nothing else moves. As long as you know index[] is sorted it works.

Note that this is almost exactly what you tried but without the extra counter. The only difference is you need to use >=0 instead of >0 in the for loop or you miss the first one.
0
 

Author Comment

by:HaniDaher
ID: 39247394
Yes, you were right I tried the one that you suggested it worked perfectly yes.
It's more simple.

Thanks.
0

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