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### 4.3 Calling the Array Example

To call the function `avg_of_double` requires making an array and then passing it as an argument. Here is an example.

```{
/* The array of values to average.  */
double nums_to_average[5];
/* The average, once we compute it.  */
double average;

/* Fill in elements of `nums_to_average`.  */

nums_to_average[0] = 58.7;
nums_to_average[1] = 5.1;
nums_to_average[2] = 7.7;
nums_to_average[3] = 105.2;
nums_to_average[4] = -3.14159;

average = avg_of_double (5, nums_to_average);

/* …now make use of `average`… */
}
```

This shows an array subscripting expression again, this time on the left side of an assignment, storing a value into an element of an array.

It also shows how to declare a local variable that is an array: `double nums_to_average[5];`. Since this declaration allocates the space for the array, it needs to know the array’s length. You can specify the length with any expression whose value is an integer, but in this declaration the length is a constant, the integer 5.

The name of the array, when used by itself as an expression, stands for the address of the array’s data, and that’s what gets passed to the function `avg_of_double` in ```avg_of_double (5, nums_to_average)```.

We can make the code easier to maintain by avoiding the need to write 5, the array length, when calling `avg_of_double`. That way, if we change the array to include more elements, we won’t have to change that call. One way to do this is with the `sizeof` operator:

```  average = avg_of_double ((sizeof (nums_to_average)
/ sizeof (nums_to_average[0])),
nums_to_average);
```

This computes the number of elements in `nums_to_average` by dividing its total size by the size of one element. See Type Size, for more details of using `sizeof`.

We don’t show in this example what happens after storing the result of `avg_of_double` in the variable `average`. Presumably more code would follow that uses that result somehow. (Why compute the average and not use it?) But that isn’t part of this topic.

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