Solved

runtime Error Erase element vector C++

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

Open in new window

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

Open in new window


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])
}

Open in new window

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;
   }

Open in new window

0
Comment
Question by:HaniDaher
  • 2
  • 2
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] <<"   ";
    }

Open in new window

0
 
LVL 37

Accepted Solution

by:
TommySzalapski earned 500 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]));
   }

Open in new window


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

Featured Post

Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Introduction This article is the first in a series of articles about the C/C++ Visual Studio Express debugger.  It provides a quick start guide in using the debugger. Part 2 focuses on additional topics in breakpoints.  Lastly, Part 3 focuses on th…
Prime numbers are natural numbers greater than 1 that have only two divisors (the number itself and 1). By “divisible” we mean dividend % divisor = 0 (% indicates MODULAR. It gives the reminder of a division operation). We’ll follow multiple approac…
The viewer will learn how to use the return statement in functions in C++. The video will also teach the user how to pass data to a function and have the function return data back for further processing.
The viewer will learn how to clear a vector as well as how to detect empty vectors in C++.

867 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

19 Experts available now in Live!

Get 1:1 Help Now