Passing Arrays as Function Arguments in c

Passing Arrays as Function Arguments in c

If you want to pass a single-dimension array as an argument in a function, you would have to declare function formal parameter in one of following three ways and all three declaration methods produce similar results because each tells the compiler that an integer pointer is going to be received. Similar way you can pass multi-dimensional array as formal parameters.

Way-1

Formal parameters as a pointer as follows. You will study what is pointer in next chapter.
void myFunction(int *param)
{
.
.
.
}

Way-2

Formal parameters as a sized array as follows:
void myFunction(int param[10])
{
.
.
.
}

Way-3

Formal parameters as an unsized array as follows:
void myFunction(int param[])
{
.
.
.
}

Example

Now consider the following function which will take an array as an argument along with another argument and based on the passed arguments, it will return average of the numbers passed through the array as follows:
double getAverage(int arr[], int size)
{
  int    i;
  double avg;
  double sum;

  for (i = 0; i < size; ++i)
  {
    sum += arr[i];
  }

  avg = sum / size;

  return avg;
}

Now let us call the above function as follows:

#include <stdio.h>

/* function declaration */
double getAverage(int arr[], int size);

int main ()
{
   /* an int array with 5 elements */
   int balance[5] = {1000, 2, 3, 17, 50};
   double avg;

   /* pass pointer to the array as an argument */
   avg = getAverage( balance, 5 ) ;

   /* output the returned value */
   printf( "Average value is: %f ", avg );
  
   return 0;
}

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

Average value is: 214.400000
As you can see, the length of the array doesn't matter as far as the function is concerned because C performs no bounds checking for the formal parameters.
Return array from function in C

C programming language does not allow to return an entire array as an argument to a function. However, You can return a pointer to an array by specifying the array's name without an index. You will study pointer in next chapter so you can skip this chapter until you understand the concept of Pointers in C.
If you want to return a single-dimension array from a function, you would have to declare a function returning a pointer as in the following example:
int * myFunction()
{
.
.
.
}
Second point to remember is that C does not advocate to return the address of a local variable to outside of the function so you would have to define the local variable as static variable.
Now consider the following function which will generate 10 random numbers and return them using an array and call this function as follows:
#include <stdio.h>

/* function to generate and return random numbers */

int * getRandom( )
{
  static int  r[10];
  int i;

  /* set the seed */
  srand( (unsigned)time( NULL ) );
  for ( i = 0; i < 10; ++i)
  {
     r[i] = rand();
     printf( "r[%d] = %d\n", i, r[i]);

  }

  return r;
}

/* main function to call above defined function */

int main ()
{
   /* a pointer to an int */
   int *p;
   int i;

   p = getRandom();
   for ( i = 0; i < 10; i++ )
   {
       printf( "*(p + %d) : %d\n", i, *(p + i));
   }

   return 0;
}

When the above code is compiled together and executed, it produces result something as follows:

r[0] = 313959809
r[1] = 1759055877
r[2] = 1113101911
r[3] = 2133832223
r[4] = 2073354073
r[5] = 167288147
r[6] = 1827471542
r[7] = 834791014
r[8] = 1901409888
r[9] = 1990469526
*(p + 0) : 313959809
*(p + 1) : 1759055877
*(p + 2) : 1113101911
*(p + 3) : 2133832223
*(p + 4) : 2073354073
*(p + 5) : 167288147
*(p + 6) : 1827471542
*(p + 7) : 834791014
*(p + 8) : 1901409888
*(p + 9) : 1990469526

It is most likely that you would not understand this chapter until you through the chapter related Pointers in C.
So assuming you have bit understanding on pointers in C programming language, let us start: An array name is a constant pointer to the first element of the array. Therefore, in the declaration:
double balance[50];
balance is a pointer to &balance[0], which is the address of the first element of the array balance. Thus, the following program fragment assigns p the address of the first element of balance:
double *p;
double balance[10];

p = balance;
It is legal to use array names as constant pointers, and vice versa. Therefore, *(balance + 4) is a legitimate way of accessing the data at balance[4].
Once you store the address of first element in p, you can access array elements using *p, *(p+1), *(p+2) and so on. Below is the example to show all the concepts discussed above:

#include <stdio.h>

int main ()
{
   /* an array with 5 elements */
   double balance[5] = {1000.0, 2.0, 3.4, 17.0, 50.0};
   double *p;
   int i;

   p = balance;

   /* output each array element's value */
   printf( "Array values using pointer\n");
   for ( i = 0; i < 5; i++ )
   {
       printf("*(p + %d) : %f\n",  i, *(p + i) );
   }

   printf( "Array values using balance as address\n");
   for ( i = 0; i < 5; i++ )
   {
       printf("*(balance + %d) : %f\n",  i, *(balance + i) );
   }

   return 0;
}
When the above code is compiled and executed, it produces following result:
Array values using pointer
*(p + 0) : 1000.000000
*(p + 1) : 2.000000
*(p + 2) : 3.400000
*(p + 3) : 17.000000
*(p + 4) : 50.000000
Array values using balance as address
*(balance + 0) : 1000.000000
*(balance + 1) : 2.000000
*(balance + 2) : 3.400000
*(balance + 3) : 17.000000
*(balance + 4) : 50.000000
In the above example p is a pointer to double which means it can store address of a variable of double type. Once we have address in p, then *p will give us value available at the address stored in p, as we have shown in the above example.
Pointers in C are easy and fun to learn. Some C programming tasks are performed more easily with pointers, and other tasks, such as dynamic memory allocation, cannot be performed without using pointers. So it becomes necessary to learn pointers to become a perfect C programmer. Let's start learning them in simple and easy steps.
As you know every variable is a memory location and every memory location has its address defined which can be accessed using ampersand (&) operator which denotes an address in memory. Consider the following example which will print the address of the variables defined:

#include <stdio.h>
int main ()
{
   int  var1;
   char var2[10];

   printf("Address of var1 variable: %x\n", &var1  );
   printf("Address of var2 variable: %x\n", &var2  );

   return 0;
}

When the above code is compiled and executed, it produces result something as follows:
Address of var1 variable: bff5a400
Address of var2 variable: bff5a3f6
So you understood what is memory address and how to access it, so base of the concept is over. Now let us see what is a pointer.

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