Invalid Iterator argument

Hi, I am trying to create a program that will take 10 numbers and then print the sum of the all two adjacent numbers and then sum of 1st and last,2nd and 2nd last;here is the code

Code:

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#include <iostream>
#include <vector>
#include <string>

using namespace std;

int main()
{
vector<int> input;

for(int i=0;i!=10;i++)
{
  int x;
  cin>>x;
  input.push_back(x);
}
for(auto it=input.begin();it!=input.end()-1;it++)
{
  int k;
  k=*it+(*(it+1));
  cout<<k<<" "<<endl;
  auto j=input.end()-it;
  cout<<*it+*j<<" "<<endl;

}

}

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.. but at the 2nd output the compiler is giving the error invalid type argument of unary '*'. where am I wrong?

Since I do not see any experienced guys answering, I am going say that auto is supported in the new standard, so if your compiler does not support it then you can not play with it. But this is just a thought..

Quote:

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Originally Posted by std10093

Since I do not see any experienced guys answering, I am going say that auto is supported in the new standard, so if your compiler does not support it then you can not play with it. But this is just a thought..

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Auto isnt the problem! my compiler is set to C11. the error messege is given when I tend to dereference j.I dont understand, if dereferencing "it" dont cause any problem.. why would "j" do..

Not an expert either but when you subtract two pointers (iterators) you are getting the distance between them, not another meaningful pointer. j is type int and not type iterator.

Quote:

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Originally Posted by Tamim Ad Dari

why would "j" do..

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j is not an iterator. when you subtract two iterators, it returns an integer result, indicating the difference between the two iterators.

so....

Quote:

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Originally Posted by Tamim Ad Dari

Auto isnt the problem!

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yes it is, at least indirectly. I'm guessing you used auto because the compiler puked on your attempt to declare it as a std::vector<int>::iterator. the trouble is that now you, as the programmer, don't really know what type j is or why it doesn't let you dereference it.

Thanks!! that helped.. I have rewrite the code. though a bit longer. And it worked.

Code:

---

#include <iostream>
#include <vector>
#include <string>

using namespace std;

int main()
{
vector<int> input;

for(int i=0;i!=10;i++)
{
  int x;
  cin>>x;
  input.push_back(x);
}
 auto j=input.end();
for(auto it=input.begin();it!=input.end()-1;it++)
{
  int k;
  k=*it+(*(it+1));
  cout<<k<<" ";

  if(j !=input.begin())
  {
      j--;
  cout<<*it+*j<<" ";
  }


}

}

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why are you using end() - 1? end() returns the iterator that is one-past the last item. the only way end() - 1 is correct, is if you do not want to process the last element.

The code uses it[1], so if it == end() - 1 (last iteration), then it[1] == end(), which is wrong.
Can't say the code looks very pretty, though.

Quote:

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Originally Posted by Elkvis

why are you using end() - 1? end() returns the iterator that is one-past the last item. the only way end() - 1 is correct, is if you do not want to process the last element.

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I am using the != syntax! not the < syntax! which means end()-1 is the last element..

Quote:

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Originally Posted by Tamim Ad Dari

I am using the != syntax! not the < syntax! which means end()-1 is the last element..

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In this case that difference is actually irrelevant.

A better solution:

Code:

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auto it = input.begin();
if (it != input.end())
{
    auto j = input.end();
    int prevValue = *it++;
    while (it != input.end())
    {
        cout << (prevValue + *it) << " ";
        --j;
        cout << *it + *j << " ";
        prevValue = *it++;
    }
}

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