The values of the variables value and list after each of the three calls to
swap are:
- value = 2 and list[5] = {1, 3, 5, 7, 9};
- value = 2 and list[5] = {3, 1, 5, 7, 9};
- value = 2 and list[5] = {3, 1, 5, 7, 9};
<h3>How to determine the values of variable value and the list?</h3>
<u>1. Passed by value</u>
In this case, the values of the actual arguments remain unchanged; So the variable value and the list would retain their initial values
The end result is:
value = 2 and list[5] = {1, 3, 5, 7, 9};
<u>2. Passed by reference</u>
With pass by reference, the arguments are changed.
Given that
value = 2 and list[5] = {1, 3, 5, 7, 9};
The value of variable value is swapped with list[0].
So, we have:
value = 1 and list[5] = {2, 3, 5, 7, 9};
Then list[0] and list[1] are swapped
So, we have:
value = 1 and list[5] = {3, 2, 5, 7, 9};
Lastly, value and list[value] are swapped
So, we have:
value = 2 and list[5] = {3, 1, 5, 7, 9};
The end result is:
value = 2 and list[5] = {3, 1, 5, 7, 9};
<u>3. Passed by value-result</u>
Since there is no loop in the program, the pass by value-result has the same output as the pass by reference.
So, the end result is:
value = 2 and list[5] = {3, 1, 5, 7, 9};
Read more about C syntax at:
brainly.com/question/15705612
