Question

Below is the code

void printLoop(type?? p){

for(int i  = 0; i<2;i++)
{
   for(int e = 0;e<3;e++)
        {
             cout<<p[i][e]<<" ";
         }
      cout<<"\n";
  }
}
void array()
{
int a[2][3] = {{1,2,3},{4,5,6}};
int (*p)[3] = a;
printLoop(p);
}

Basic idea is that I want to print out the array using a for loop in the printLoop func. However, I need to know the type of that pointer which has the address of the 2D array. What’s the pointer’s type? Is it int (*)[]? I’m confused.

Also what does “(*p)” mean(from int (*p)[3]) ? Thanks a lot!

Answer 1

what does “(*p)” mean(from int (*p)[3]) ?

p is a pointer to an array of size 3 of objects of type int.

You have multiple possibilites for your printLoop function (though with the general C-restriction that you can leave at most one — the outermost declarator empty):

• You can specify the dimensions explicitly:

  void printLoop(int p[ 2 ][ 3 ]);

The only advantage with this method is that the implementation can consider that the array being passed is of the desired size (i.e. 2×3 matrix of ints) as a pre-condition.

• You can leave out the [ 2 ] part entirely:

  void printLoop(int p[][ 3 ]);

or,

void printLoop(int (*p)[ 3 ]);

• You can use a pointer to a pointer of int

You will also need to pass the dimensions (if you skip one that is) along to make sure that you don’t access out-of-bounds memory. So, your function signature should go like this:

void printLoop(int (*p)[ 3 ], int dim);