Modifying a pointer to an array in a function

Question

I have a problem with passing an array to a function and then modifying it. For example:

void foo(int ** a) {
  int * b = new int[3]
  //Initialize b, i.e b = {3, 2, 1}
  a = &b;
  //*a = {3, 2, 1}
}

int * c = new int[3]
//Initialize c; c = {1, 2, 3}
foo(&c);
// c is still {1, 2, 3}. Why?

I'm not really sure why c doesn't point to the same array as b.

*a = b; Should work. Though I am not sure what you are trying to do. — Arunmu
– Arunmu, Commented Jul 5, 2016 at 13:39
@MarkoStojanovic Your question is little similar to the behavior explained in this question so it will also help you understand your problem :) — Itban Saeed
– Itban Saeed, Commented Jul 5, 2016 at 14:00
@ItbanSaeed Yes, I was looking for exactly that, but didn't find it myself. I guess I didn't know how to form the statement properly when googling. — Marko Stojanovic
– Marko Stojanovic, Commented Jul 5, 2016 at 14:07
its no problem, there is no rocket science in googling. You might have not reached the link from the results.. I just thought about making your issue little more clear :) — Itban Saeed
– Itban Saeed, Commented Jul 5, 2016 at 14:13

Mattia F. · Accepted Answer · 2016-07-05 13:41:57Z

5

a has type int** and it's passed by value, so modifying it does not modify c.

To modify it, you have to do this:

  *a = b;

By doing this, you assign the address of b to the variable pointed by *a (which corresponds to c), so they will point to the same array.

answered Jul 5, 2016 at 13:41

Mattia F.

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1 Comment

Marko Stojanovic Over a year ago

Thank you. I forgot that c's location is passed by value, so I thought *a = b and a = &b would be equivalent.

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