Make array in cpp

Question

I am trying to make a new array in my project

the code is:

#include <iostream>
using namespace std;
void makeArray( int *& arrayPtr, int size );
int main( )
{
  int * arrPtr;
  int size =10;
  makeArray( arrPtr, size );
  for(int j=0;j<size;j++)
  {
    cout<<arrPtr[j]<<endl;
  }
}
void makeArray( int *& arrayPtr, int size )
{
  int arr[size-1];
  for(int i=0;i<size;i++)
  {
    arr[i]=0;
  }

  *&arrayPtr=*&arr;
}

According to the requirements i need to use the above "makeArray" method inorder to make the array. When i run the code the output is garbage values not zero.....

any help will be appreciated

thank you

I think the last line is a bit awkward: having *& is the same as removing it, i.e. the line would be equal to *&arrayPtr=arr; — Michel Keijzers
– Michel Keijzers, Commented Feb 12, 2012 at 2:56
So why are you doing this when there's a perfectly good std::vector class? Is this schoolwork? — Ben Voigt
– Ben Voigt, Commented Feb 12, 2012 at 2:58

Michael Chinen · Accepted Answer · 2012-02-12 03:07:00Z

4

The way you are creating the array is on the stack, which means that it will not exist after the makeArray function finishes.

You will need to allocate the array on the heap.

So:

int arr[size-1];

should be:

int *arr = new int[size-1];

Also, I think you mean to do this in makeArray():

arrayPtr = arr;

Instead of:

*&arrayPtr=*&arr;

Which compiles but is more complex and is functionally the same thing in this context.

But you may prefer just returning an int* instead of taking a reference to the pointer.

Then when you are done using the array in main(), and set it to NULL just in case you accidentally use it again, like this:

for(int j=0;j<size;j++)
{
  cout<<arrPtr[j]<<endl;
}
delete [] arrPtr;
arrPtr = NULL;

edited Feb 12, 2012 at 3:07

answered Feb 12, 2012 at 2:55

Michael Chinen

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1 Comment

johnsyweb Over a year ago

You might like to highlight where you'd delete[] this to avoid memory leaks.

pg1989 · Accepted Answer · 2012-02-12 02:58:45Z

0

Why are you declaring a parameter as 'int *& arrayPtr'? Do you just need a pointer to an array? You should use 'int *arrayPtr' instead.

To answer your question, the problem is that you are declaring an array in the function makeArray's stack. Upon the completion of a function, that function's stack is destroyed, so you're passing the address of junk data. To avoid this, use dynamic memory allocation instead.

EDIT: Also, you should use memset instead of a for loop to zero an array. It's much faster.

answered Feb 12, 2012 at 2:58

pg1989

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2 Comments

Ben Voigt Over a year ago

Because he's reseating the pointer, he does need to pass it by reference.

pg1989 Over a year ago

Oh, ok. My C++-fu is a little rusty.

uriel · Accepted Answer · 2012-02-12 03:03:49Z

0

The "arr" which you allocate in "makeArray()" is local. and when the functione is over the array is release. When you back to main you get garbage. What you want to do, is to use the "new" operator to allocate this new array to be used in all program, unless you will free this memory by "delete". so you can set your makeArray() to:

int* makeArray(int size )
{
  int *arr = new[size];
  for(int i=0;i<size;i++)
  {
    arr[i]=0;
  }

  return arr;
}

the in you main you need to initialize your arry by:

int * arrPtr = makeArray(10);

just don't forget to release this memory after you finsh:

delete[] arrPtr ;

answered Feb 12, 2012 at 3:03

uriel

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