A function that takes an int arg and returns a pointer to an int[5] array

[SSteven]

Hi

I am trying to create a simple function that takes an int arg and returns a pointer to an int[5] array.

The array and function definition are defined as follows and compile correctly.
However, the code to invoke the function and receive the returned array, won't compile.

   int ia[5] = {10, 20, 30, 40, 50};   /// array

   int (* f3a (int)) [5]               /// function
   {
      return &ia;
   }

   int main()
   {
      
      int* iar1 = f3a(1);

      return 0;
   }

This gives the following compilation error:

error: invalid conversion from 'int (*)[5]' to 'int*' [-fpermissive]|


To solve this problem, I need to define the function differently - to return an int pointer, and then the function code returns the array. Then, invocation is perfectly OK.

The question is : How can the first technique to define a function that returns an array, be made to work? It is based on an example given by Bruce Eckel in TICPP. Hence the query.


Thanks
Steven

[keskiverto]

The error says that the type returned by f3a is not int* (or anything that would implicitly convert to int*).

One can get around that error:

int main() {
  auto iar1 = f3a(1);
  return 0;
}

However, using iar1 might turn out to be difficult.


What are you trying to achieve (aka I have no idea what the TICPP is)?

[hiphop12ism]

One can get around that error:
int main() {
auto iar1 = f3a(1);
return 0;
}

Unfortunately, this never worked for me
Just like keskiverto, i also dont have any idea about TICPP.
But this is how it worked:

 int ia[5] = {10, 20, 30, 40, 50};   /// array

   int (* f3a (int)) [5]               /// function
   {
      return &ia;
   }

   int main()
   {
      int (* f3as ) [5]  = f3a(1);
      return 0;
   }

if you want to print the values of the array.

This works:

for(int i=0; i<5; i++){
        cout<<*(f3as[0]+i)<<endl;
      }

10
20
30
40
50

[keskiverto]

The auto requires C++11.

The *(f3as[0]+i) is same as f3as[0][i]

The int (*foo) [5] is a "pointer to array". It is just a pointer that returns a pointer, but it has special pointer math coefficient.

If bar is an int array of some size and foo points to its first element &(bar[0]) == &(foo[0][0]), then &(foo[x][y]) == &(bar[5*x+y])

The pointer to array syntax is rarely used. As you should notice from above, it allows using (via pointer) an 1D array like it were a 2D array.

[SSteven]

Thanks for the replies. I shall study them in detail.

TICPP stands for "Thinking in C++", the free e-book written by Bruce Eckel, who is also famous for his book TIJ ("Thinking in Java"). Both e-books are freely downloadable from Bruce Eckel's website www.bruceeckel.com.

Steven

[SSteven]

Hi

Yes, The following code is working and thanks very much for the same:

int ia[5] = {10, 20, 30, 40, 50};   /// array

int (* f3a (int)) [5]                      /// function
{
   return &ia;
}

int main()
{
    /// Invoke the function directly.
    int (*iar1) [5]  = f3a(1);	             /// WORKS PERFECTLY
    print(*iar1);

    return 0;
}

However, I have also tried defining a pointer to the above function. That works's fine, but when I try invoking the function through the pointer, the program crashes:

int (* (*fp3a) (int)) [5];                  /// function pointer

/// Invoke the function through the pointer.
int (*iar2) [5] = (*fp3a) (1);	    /// PROGRAM CRASHES AT THIS LINE 

Any solution?

Thanks
Steven

[keskiverto]

Are you dereferencing an uninitialized pointer?

[SSteven]

Yes, you are quite right. Actually, I had forgotten to initialize the function pointer to the function's address, and was trying the dereference the pointer though it was uninitialized.

The corrected code works perfectly:

#include <iostream>
#include <exception>

int ia[5] = {10, 20, 30, 40, 50};   /// array

int (* f3a (int)) [5]                      /// function
{
   return &ia;
}

int (* (*fp3a) (int)) [5] = f3a;     /// function pointer

/// Helper function to print an array.
void print(int ia[])
{
    for(int i = 0; i < 5; i++)
        cout << ia[i] << " ";
    cout << endl;
}

/** Arrays to store Function return values
     Each variable is a pointer to an int[5] array.

     This is necessary since the function returns
     a pointer to an int[5] array.
*/
int (*iar1) [5];
int (*iar2) [5];

int main()
{
    /// Invoke the function directly.
    iar1  = f3a(1);
    print(*iar1);

    /// Invoke the function through the pointer.
    iar2  = (*fp3a)(1);
    print(*iar2);

    return 0;
}

Thanks.
Steven