Array resizing using reference / pointers

Question

I made a small code to resize a dynamic array, passing pointer X as a reference.

void resize(int*& X , int & dimX){
int * new_X = new int [dimX+20];
 for(int i=0;i<dimX;i++)
    new_X[i] = X[i];
 delete [] X;
 X = new_X;
 dimX += 20;
}

My doubt is: what would be the difference in the code if i decided to pass the array X as a sole pointer? For example:

void resize(int* X , int & dimX)

Is that even possible for this kind of operation? (resizing). Thanks a lot and sorry for the dumb question, i'm a beginner.

Raw arrays are not "dynamic" (as you can see with the work you need to do!). Use std::vector or std::list if you need a "dynamic" collection. — crashmstr
– crashmstr, Commented Jul 9, 2015 at 15:28
@crashmstr Thanks for info :) i know about the vector class, but i just need to do a resize function by myself, and what i was wondering is: what's the best way to pass the array X to the function? and how passing int * X instead of int *&x would affect the code. — Mastarius
– Mastarius, Commented Jul 9, 2015 at 15:30
To modify a something which is passed as argument to a function you need to pass its reference or a pointer to that thing. So if you want to modify a pointer, you need to pass a reference to a pointer. — UmNyobe
– UmNyobe, Commented Jul 9, 2015 at 15:31

MahlerFive · Accepted Answer · 2015-07-09 15:34:59Z

1

If you pass X as an int*, then you are passing a copy of the pointer. This means if you change X on the line X = new_X;, you will only update the copy, not the original.

You can either keep using a reference to the pointer like you are currently doing, or take X as an int* but return new_X and have the caller use the returned value.

answered Jul 9, 2015 at 15:34

MahlerFive

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