CODEWITHSUNDEEP
c++
user103052
user103052

Reputation: 51

Pointers and dynamically allocated arrays in C++

I'm trying to get this clear up in my mind.

int* arrayA = new int[3];
int arrayB[3] = {1,2,3}

arrayA = arrayB;

Does the value of arrayB get copied over to arrayA? Or does arrayA become a pointer to arrayB?

Upvotes: 5

Views: 6371

Answers (5)

AnotherDeveloper
AnotherDeveloper

Reputation: 2201

Q. Does the value of arrayB get copied over to arrayA? 
Ans. NO

Or

Q. does arrayA become a pointer to arrayB?
Ans YES.

But the memory chunk which you allocated using new is lost. As no one is having the address which was starting point of that chunk.

This is vindicated by this program :

#include<iostream>
using namespace std;

const int MAX=4;
int main(){
   int* arrayA = new int[MAX];
   for(int i=0;i<MAX;i++){  arrayA[i]=i;    }
   int arrayB[MAX] = {10,20,30,40};
   arrayA = arrayB;
   for(int i=0; i<MAX;i++){ cout<<" a["<<i<<"] = "<<arrayA[i]<<endl;    }
   return 0;
}

The output of this code is :

 a[0] = 10
 a[1] = 20
 a[2] = 30
 a[3] = 40

on

g++ (Ubuntu 4.8.4-2ubuntu1~14.04) 4.8.4

which is true in general as well.

visualization of this is somewhat like this

arrayA-> [0][1][2][3]
arrayB-> [10][20][30][40]

Intially arrayA has address of [0] ( + sum extra memory on the negative indexes for size used by compiler to do the de-alloctation ).

after
arrayA=arrayB;

arrayA has address of [10];

suppose you call delete on arrayA what shall happen ??

*** Error in `./a.out': double free or corruption (out): 0x00007fff561281d0 ***

This is evident from the following code also

#include<iostream>
using namespace std;

const int MAX=4;
int main(){
    int* arrayA = new int[MAX*2];
    for(int i=0;i<MAX;i++){ arrayA[i]=i;    }
    int* arrayC = new int[MAX];
    int j=0;
    for(int i=MAX*100;i>=0;i-=10,j++){  arrayC[j]=i;    }
    arrayA = arrayC;
    for(int i=0; i<MAX;i++){    cout<<" a["<<i<<"] = "<<arrayA[i]<<endl;    }

    delete[] arrayA;
return 0;
}

The out put is : 

 a[0] = 400
 a[1] = 390
 a[2] = 380
 a[3] = 370
*** Error in `./a.out': free(): invalid next size (fast): 0x0000000002028040 ***
Aborted (core dumped)

Just pay attention to :

"invalid next size"

This clearly says about the size value written before the arrayA in the negative indexes.

Upvotes: 0

Pradeep Kumar
Pradeep Kumar

Reputation: 65

Array name represents the base address of the array.

arrayA=arrayB;

stores the base address of arrayB to arrayA. The further values can be accessed by incrementing the pointer arrayA.

i.e

*(arrayA) value is 1

*(arrayA+1) value is 2

*(arrayA+2) value is 3

The memory created should be freed at the last using

delete[] arrayA

Upvotes: 0

Jan Hudec
Jan Hudec

Reputation: 76246

Naked new is evil, especially for arrays. Use std::vector if you need dynamic array and std::array if you need static one.1

Now the actual answer is simple. arrayA is a pointer. So it's made to point to arrayB. That has additional effect that:

  1. you now leaked the allocated pointer.
  2. attempt to delete arrayA will now crash or something else unpleasant, because you would be trying to free unallocated memory.

Actually while C++ allows assigning structures and classes, it does not support assigning arrays at all. Unless part of an object either managed as in std::vector or as array member as in std::array (note: std::array is C++11).

1 The main reason is that the containers guarantee the resources will be released in their destructors, so called RAII idiom, avoiding most opportunities for memory leaks and dangling references.

Upvotes: 7

Saravanan
Saravanan

Reputation: 1440

For

Does the value of arrayB get copied over to arrayA?

Nope. It will not copy arrayB to arrayA. If you want to do so, you can do like,

for (int i=0;i<3;i++) arrayA[i]=arrayB[i]; //This is two arrays has same value

does arrayA become a pointer to arrayB?

Yes. It become a pointer to array of arrayB. After arrayA = arrayB; statement, memory allocated in the first line can not be accessible and it is a leak.

arrayA=arrayB; result in copying the address of arrayB to arrayA. arrayA will start pointing to the arrayB. Check it using,

printf("\n Base address of A[%u] base address of B[%u]", arrayA, arrayB);

Hope this helps.

Upvotes: 2

Michael Shaw
Michael Shaw

Reputation: 644

int* arrayA = new int[3];

arrayA now stores the address in memory of a different memory block thats large enough to contain 3 ints which is stored on the heap.

int arrayB[3] = {1,2,3};

arrayB is a block of memory thats large enough to contain 3 ints which is stored on the stack.

arrayA = arrayB;

Copies the address of the memory block that's stored on the stack into the variable arrayA. You have now lost the only reference to memory block stored on the heap, and have no way to free up the memory (a memory leak)

The memory stored on the stack and now pointed to by arrayA will not be safe to access when the current function returns. In particular returning arrayA (or arrayB) is not safe.

Upvotes: 6

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