C++ Returning Pointer from Functions

C++ Pointers

In the previous chapter, we learned how to return an array from a function in C++. Similarly, C++ allows you to return a pointer from a function. To do this, you must declare a function that returns a pointer, as follows:

int * myFunction()
{
.
.
.
}

Additionally, C++ does not support returning the address of a local variable outside the function, unless the local variable is defined as a static variable.

Now, let's look at the following function, which generates 10 random numbers and returns them using an array name that represents a pointer (i.e., the address of the first array element), as follows:

Example

#include <iostream>
#include <ctime>
#include <cstdlib>
 
using namespace std;
 
// function to generate and return random numbers
int * getRandom( )
{
  static int  r[10];
 
  // Set seed
  srand( (unsigned)time( NULL ) );
  for (int i = 0; i < 10; ++i)
  {
    r[i] = rand();
    cout << r[i] << endl;
  }
 
  return r;
}
 
// main function to call the function defined above
int main ()
{
   // a pointer to an integer
   int *p;
 
   p = getRandom();
   for ( int i = 0; i < 10; i++ )
   {
       cout << "*(p + " << i << ") : ";
       cout << *(p + i) << endl;
   }
 
   return 0;
}

When the above code is compiled and executed, it produces the following result:

624723190
1468735695
807113585
976495677
613357504
1377296355
1530315259
1778906708
1820354158
667126415
*(p + 0) : 624723190
*(p + 1) : 1468735695
*(p + 2) : 807113585
*(p + 3) : 976495677
*(p + 4) : 613357504
*(p + 5) : 1377296355
*(p + 6) : 1530315259
*(p + 7) : 1778906708
*(p + 8) : 1820354158
*(p + 9) : 667126415

C++ Pointers

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