CODEWITHSUNDEEP
c++
Missy Zewdie
Missy Zewdie

Reputation: 231

Heap Corruption Detected in Dynamic Array c++?

I was learning how to use the delete operator for dynamic arrays and i ran into some really strange problem. I tried two approach to write the same code - one worked but the other dint work. So, here is the code i tried

Method 1:- 

#include <iostream>
using namespace std;

int main()
{

  int *p = new int(2);

   for(int i =0; i<2;i++)
   {
       cin>>p[i];

   }

   for(int i=0; i<2;i++)
   {
       cout<<p[i];

   }

      delete[] p;

     system("pause");
}

//Error message - Heap Corruption detected. Can someone please explain me the reason of the error?

Method 2:- Using TypeDef 

#include <iostream>
using namespace std;

typedef int arr[2];

int main()
{

  int *p = new arr;

   for(int i =0; i<2;i++)
   {
       cin>>p[i];

   }

   for(int i=0; i<2;i++)
   {
       cout<<p[i];

   }

      delete[] p;

     system("pause");
}

//The above method works perfectly with no error. I am totally confused!!

Upvotes: 0

Views: 318

Answers (4)

Emil Laine
Emil Laine

Reputation: 42828

new int(2) allocates one int that has the value 2.

new int[2] allocates an array of two ints, with indeterminate values.

Upvotes: 2

Nick
Nick

Reputation: 24

in your first example you have:

int *p = new int(2);

which is initializing 1 pointer to the value of 2 rather than in your second example when you have a

new int[2];

where you're actually making 2 pointers.

Upvotes: 0

Austin Brunkhorst
Austin Brunkhorst

Reputation: 21130

You're calling the constructor of one int with the value of 2 not allocating two ints.

Use square brackets instead.

int *p = new int[ 2 ];

Upvotes: 2

Bathsheba
Bathsheba

Reputation: 234665

First case: you are only creating one int and assigning it the value of 2.

Then you treat it as an array with two elements. That's undefined behaviour. You need to use new int[2]. Your delete[] syntax is correct.

Your second example is far too obfuscated.

Upvotes: 2

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