Reverse array pointer in c++

Question

I am trying to learn pointers with arrays.
I want to do this with the function below:

int* ArrayManipulator(int* arrayPoiner, const int size)

to reverse the array list:

   #include <iostream>

using namespace std;

int* ArrayManipulator(int* arrayPoiner, const int size)
{
    int i=0;
    int tt[size];
    for (i=0;i<=size;i++)
    {
        cout<<arrayPoiner[size-i]<<endl;
        cout<<&arrayPoiner[size-i]<<endl;

        tt[i]=arrayPoiner[size-i];
    }
    return tt;
}


int main()
{
    int t[]={1,2};
    int *ReversedArray;
    ReversedArray = ArrayManipulator(t,1);
    int i=0;
    for (i=0;i<2;i++)
    {
        cout<< ReversedArray[i]<<endl;
    }
  return 0;
}

I am receiving error at cout<< ReversedArray[i]; cout<< &ReversedArray[i]; is printing the address of the value but I am unable to print the value - codeblocks error out there

EDIT Code:

I want the final two lines to be 2 and 1, where am I going wrong on code.

you should get SEGV , and tt should be on the heap, and you also forgot to return the pointer. — g24l
– g24l, Commented Dec 8, 2015 at 23:33
ArrayManipulator returns a pointer to an array that was allocated on the stack; this invokes undefined behaviour. — Jonathan Potter
– Jonathan Potter, Commented Dec 8, 2015 at 23:38
You are also passing the size of the array - 1 to your function and then declaring int tt[size] or tt[1] so that tt has only 1 element — Bob__
– Bob__, Commented Dec 8, 2015 at 23:58

g24l · Accepted Answer · 2015-12-08 23:47:39Z

1

You must return the pointer to the array that has been allocated on the heap:

Modify your method signature:

int** ArrayManipulator(int* arrayPoiner, const int size)

Allocate on the heap:

int **tt = new int*[size];

return the pointer to the array;

return tt;

and modify your calls such that:

int **ReversedArray = ArrayManipulator(t,1);

and to output do this:

cout<< *ReversedArray[i];

Don't forget to delete the space allocated with:

  delete [] ReversedArray;

edited Dec 8, 2015 at 23:47

answered Dec 8, 2015 at 23:37

g24l

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4 Comments

surpavan Over a year ago

I was beating my head for this - now need to learn more on ** pointers :(

surpavan Over a year ago

Thank you very much for the time and knowledge.

Bob__ Over a year ago

Why are you using a **?

g24l Over a year ago

@Bob__ the OP modified the question a bit since this was answered. In the original version he was actually using a array to pointers.

Vlad from Moscow · Accepted Answer · 2015-12-09 00:04:41Z

1

Function ArrayManipulator returns nothing. So the program is already has undefined behaviour.

I think what you want to get is the following

#include <iostream>

using namespace std;

int * ArrayManipulator( const int *source, size_t n, int *dest )
{
    dest = dest + n;

    while ( n-- ) *--dest = *source++;

    return dest;        
}

int main()
{
    const size_t N = 2;

    int a[N] = { 1, 2 };
    int b[N];

    int * ReversedArray = ArrayManipulator( a, N, b );

    for ( size_t i = 0; i < N; i++ )
    {
        cout << ReversedArray[i] << ' ';
    }

    cout << endl;

    return 0;
}

Take into account that there is standard algorithm std::reverse_copy declared in header <algorithm> that does a similar work. If you want to reverse an array "in place" then you can use another standard algorithm std::reverse.

edited Dec 9, 2015 at 0:04

answered Dec 8, 2015 at 23:42

Vlad from Moscow

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1 Comment

g24l Over a year ago

that is actually a very nice solution to this problem, if one wants to avoid heap allocation.

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