[5j03] Passing Arrays as Function Arguments in C

https://www.tutorialspoint.com/cprogramming/c_passing_arrays_to_functions.htm

If you want to pass a single-dimension array as an argument in a function, you would have to declare a 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. Similarly, you can pass multi-dimensional arrays as formal parameters.

Way-1

Formal parameters as a pointer −

void myFunction(int *param) {
   .
   .
   .
}

Way-2

Formal parameters as a sized array −

void myFunction(int param[10]) {
   .
   .
   .
}

Way-3

Formal parameters as an unsized array −

void myFunction(int param[]) {
   .
   .
   .
}

Example

Now, consider the following function, which takes an array as an argument along with another argument and based on the passed arguments, it returns the average of the numbers passed through the array as follows −

double getAverage(int arr[], int size) {

   int i;
   double avg;
   double sum = 0;

   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 the 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 formal parameters.