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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.

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4
  • See stackoverflow.com/questions/29735567/…. Different language, same problem. Your code looks like C anyway. Commented Apr 20, 2015 at 5:23
  • don't bother using those language supported arrays. even the more because you declare them on the stack, you are going to cause stackoverflows. just use some nice matrix classes, gee. Commented Apr 20, 2015 at 5:23
  • See stackoverflow.com/questions/3911400/passing-2d-arrays to understand how to pass 2D arrays. Hopefully that will help you understand how to pass 2D arrays. Commented Apr 20, 2015 at 5:32
  • By the way: You are accessing your arrays out of bounds in line 11 and 12. Declaring an array with size 2 and accessing it with ary[2] is undefined behavior. The index to access the array is 0-based, that means that to access the second element you have to write ary[1]. Commented Apr 20, 2015 at 7:13

4 Answers 4

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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];.

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Comments

1

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
           }
       }
   }
}
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0

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];
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3 Comments

got it! Thanks. Edited answer accordingly :)
I'm sorry, how is ary[5][10] referring to 1D?
Image you are in a room, which has height, width and length. If you give length and width then the remaining is height, which is 1D, isn't it? :)
0

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);
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