passing 1d and 2d arrays by reference in c++

Question

I am new to c++ and have a couple questions regarding passing arrays by reference to functions (so that the arrays are modified by the function). I realize there are similar questions that have been asked already, but there are a few points that I think were not covered in those previous questions (at least from what I saw). From what I have gathered so far, one can pass an array by reference by doing the following:

#include<iostream>
using namespace std;

void modify_array(int* a);

int main()
{
  int array[10];

  modify_array(&array[0]);

  for(int i=0;i<10;i++)
  {
    cout<<array[i]<<endl;
  }
}

void modify_array(int* a)
{
  int i;
  for(i=0;i<10;i++)
  {
    *(a+i)=i;
  }
}

This makes sense to me but if I change the function to:

void modify_array(int* a)
{
  int i;
  for(i=0;i<10;i++)
  {
    a[i]=i;  //line changed
  }
}

This also works. Is there a difference? Or is the second just a short cut? Also in the case of passing 2d arrays I would have guessed that the following code would work:

#include<iostream>
using namespace std;

void modify_array(int* a);

int main()
{
  int array[10][10];

  modify_array(&array[0][0]);

}

void modify_array(int* a)
{
  int i,j;
  for(i=0;i<10;i++)
  {
    for(j=0;j<10;j++)
    {
      a[i][j]=i*j;
    }
  }
}

But this doesn't. From what I have seen in other related questions, you would do something like:

void modify_array(int (*a)[10])
{ 
  int i,j;
  //a[i][j]= blah blah blah;
}

or,

void modify_array(int (&a)[10][10])
{ 
  int i,j;
  //a[i][j]= blah blah blah;
}

What is the difference between these latter two function definitions? What do experienced c++ programmers recommend using: the (*a)[10][10] notation or the (&a)[10][10] notation?

Answer 1

Writing *(a+i)=i; or a[i]=i; are equivalent. The first is seen as an offset applied to the pointer to the array and assigning the value to the pointee, while the second is assigning the value to the element of the array.

However, when passing a pointer to a function modify_array(int* a), it cannot deduce that the pointee is a 2D array, and does not know what size to offset to address the other lines of the array, with a[i][j]=i*j;. For the compiler, it can only access the first dimension of the array.

The proper way to do what you need is this

#include<iostream>
using namespace std;

void modify_array(int (&a)[10][10]);

int main()
{
  int array[10][10];

  modify_array(array);

  for(int i=0;i<10;i++)
  {
    for(int j=0;j<10;j++)
    {
      cout<<array[i][j]<<endl;
    }
  }
}

void modify_array(int (&a)[10][10])
{
  int i;
  for(i=0;i<10;i++)
  {
    for(int j=0;j<10;j++)
    {
      a[i][j]=i*j;
    }
  }
}

Live example

The function is expecting an int array of 10x10, passed by reference.