Passing arrays as pointer parameters - c++

Question

I was trying to dig more into using arrays and pointers while I came across this problem:
main.cpp:13:7: error: cannot convert 'int [2][5][10]' to 'int*'
in assignment show=ary;
Here's the code:
#include <iostream>
using namespace std;
void arrayLearn(int *);
int main()
{
int ary[2][5][10];
int *show;
ary[2][4][9]=263;
ary[2][5][10]=100;
show=ary; //line 13
arrayLearn(show); //line 14
cout <<"End process"<<endl;
return 0;
}
void arrayLearn(int *arg)
{
for(int i=0;i<100;i++)
{
cout<<"Pointer position: "<<i<<" Value: "<<*(arg++)<<endl;
}
}
If I remove the line 13 and replace line 14 with the following code,
arrayLearn(ary[5][10]);
then the program compiles, but I don't understand why I should pass only two dimensions and not three. If I'm passing the pointer to the first item in the array then why can't I just pass the pointer like this?
arrayLearn(ary);
Please let me know if I've missed some vital concepts or failed to see a really simple mistake.

Answer 1

As Abhishek stated, you are trying to pass a 3-dimensional array to a 1-dimensional object.
If the show variable is not used anywhere else, you can do the following:
#include <iostream>
using namespace std;
void arrayLearn(int *);
int main()
{
int ary[2][5][10];
// Code change explained below (see Note)
ary[1][4][8] = 263;
ary[1][4][9] = 100;
// Pass the starting address of the array
arrayLearn(&ary[0][0][0]);
cout <<"End process"<<endl;
return 0;
}
// Iterate through the array using the position of the elements in memory
void arrayLearn(int *arg)
{
for(int i=0;i<=100;i++)
{
cout<<"Pointer position: "<<i<<" Value: "<<*(arg++)<<endl;
}
}
Output:
...
Pointer position: 98 value: 263
Pointer position: 99 value: 100
End process
This way allows you to pass the starting address of the array to the function.
Note: It should be noted that your original array assignments ary[2][4][9] = 263; and ary[2][5][10] = 100; are outside the bounds of the array. Array indexes start at 0. So even though you have declared your array as ary[2][5][10]; To access the last array element, you would use ary[1][4][9];.

Answer 2

If you are willing to use std::vector, which I highly recommend, you can use the following code to create a 2D vector:
using std::vector<int> Vec1;
using std::vector<Vec1> Vec2;
using std::vector<Vec2> Vec3;
Vec3 a(2, Vec2(5, Vec1(10, 0));
and then change the argument of arrayLearn to const Vec3& to call it using a.
void arrayLearn(const Vec3& arg)
{
// You'll need to use three loops to traverse all the elements.
for ( auto const& v2 : arg )
{
for ( auto const& v1 : v2 )
{
for ( auto const& item : v1 )
{
// Use item
}
}
}
}

Answer 3

In simple terms, ary is 3-Dimensional and show is 1-Dimensional. The compilation errors tell you that you can't convert 3D into 1D. Now, why does following work?
arrayLearn(ary[5][10]);
It is because, ary[5][10] is referring to 1D (a row or a column or a height, depends on how you visualize 3D array) and arrayLearn is also accepting 1D parameter.
if you want to pass ary using show then pass 2D like ary[5][10] to show variable, something like following :)
show = ary[5][10];

Answer 4

you can use STL data structurs like vector, map , linkedlist , ...
that all of them are better that array both in emplementation and in performance.
but for solving exactly this problem, you must pass 3 dimention array to one function in this manner:
int array[10][7][8];
void function_sample(int (*a)[7][8] , int length)
{
//for example for printing array cells
for(int i =0 ; i< length ; i++)
{
for(int j =0 ; j< 7 ; j++)
{
for(int k =0 ; k< 8 ; k++)
{
cout << a[i][j][k] << endl
}
}
}
//....
}
and for calling this function :
int array[10][7][8] //= {initial your array};
function_sample(array , 10);