Question

include using namespace std; void tryMe(int& v); int main() { int x = 8; for (int count = 1; count < 5; count++)… # What is the output of the following program? #include using namespace std; void tryMe(int& v); int main() { int x = 8; for (int count = 1; count < 5; count++) tryMe(x); return 0; } void tryMe(int& v) { static int num = 2; if (v % 2 == 0) { num++; v = v + 3; } else { num--; v = v + 5; } cout << v << ", " << num << endl; }

Short answer

The output is: '11, 3', '16, 2', '19, 3', '24, 2'.

Step-by-step solution

Initial Setup in main() Function

We start in the main function with the initialization of the integer variable \( x \) set to 8. Then, a for-loop begins with an integer variable \( count \) starting from 1, looping as long as \( count \) is less than 5, incrementing by 1 each loop. Inside the loop, the function \( tryMe(x) \) is called.

Understanding the tryMe() Function

The function \( tryMe() \) takes an integer reference variable \( v \). It declares a static integer \( num \) initialized to 2 the first time the function is called. Since \( num \) is static, it retains its value between function calls.

First Loop Iteration - count = 1

For \( count = 1 \), the function \( tryMe(x) \) is called with \( x = 8 \). Since 8 is even, \( num \) is incremented to 3 and \( x \) is increased by 3, making \( x = 11 \). The output is '11, 3'.

Second Loop Iteration - count = 2

For \( count = 2 \), \( tryMe(x) \) is called with the updated \( x = 11 \). Since 11 is odd, \( num \) is decremented to 2 and \( x \) is increased by 5, making \( x = 16 \). The output is '16, 2'.

Third Loop Iteration - count = 3

For \( count = 3 \), \( tryMe(x) \) is called with \( x = 16 \). Since 16 is even, \( num \) is incremented to 3 and \( x \) is increased by 3, making \( x = 19 \). The output is '19, 3'.

Fourth Loop Iteration - count = 4

For \( count = 4 \), \( tryMe(x) \) is called with \( x = 19 \). Since 19 is odd, \( num \) is decremented to 2 and \( x \) is increased by 5, making \( x = 24 \). The output is '24, 2'.

Conclusion of the Loop

The loop ends as \( count \) becomes 5, which is not less than 5, thus exiting the loop and ending the program.

Key concepts

Static Variables

Static variables in C++ hold a special place, especially when you need a certain value to persist between function calls. As shown in the exercise, the variable `num` inside the `tryMe` function is initialized as a static integer. This means that `num` is only initialized once and retains its value for the next time the function is called.

A static variable is declared using the `static` keyword. Unlike regular local variables, which get re-initialized every time the function is called, a static variable gets initialized once and keeps its value even after the function exits. This is extremely useful when you want to maintain state or count something across multiple invocations of a function.

In our example, the static variable `num` starts at 2, and with each call to `tryMe`, it remembers whether to increment or decrement based on the operations performed during the function's previous executions. This characteristic allows it to "remember" the condition of the previous call, illustrating its persistence.

Loop Iteration

Loops are a foundational aspect of C++ programming, used widely to perform repetitive tasks efficiently. The for-loop in our problem is straightforward but powerful, executing the `tryMe(x)` function four times in succession.

The loop begins with the declaration of an integer `count`, initialized to 1. The loop continues to iterate as long as `count` is less than 5, and increments `count` by 1 after each completion of the loop body. This means the code inside the loop will execute for `count` values of 1, 2, 3, and 4.

For-loop is particularly advantageous in providing a clear syntax to control iteration using initialization, condition, and increment expressions. Thus, allowing you to manage how many times a code block should run precisely, which in this case controls how many times `tryMe(x)` is called.

Function Calls

A function call in C++ involves invoking a function to perform a specific operation. In the given program, the `tryMe(x)` function call is integral to the solution, as it processes the logic affecting the variable `x` and modifies the static variable `num`.

Function calls encourage code reuse and modularity, meaning you can use the same function multiple times throughout your program without rewriting code. Each call to `tryMe(x)` passes the current value of `x` as a parameter. The function executes its logic, performs calculations, and potentially changes the state of its parameters, as well as static and global variables.

In simple terms, every time `tryMe` is called, it performs calculations independently of its previous calls except for the changes in `num`, thanks to its static nature. This feature aids in maintaining streamlined and clean code, especially for tasks that are logically grouped under a specific function.

Reference Parameters

In C++, reference parameters provide a way for a function to modify the argument passed to it. By using reference parameters, you can directly affect the original variable rather than just a copy of it.

The reference parameter is denoted by an `&` symbol after the data type in the function's definition. In the `tryMe` function, `int& v` signifies that `v` is a reference to the argument passed during the function call. This means any changes applied to `v` within `tryMe` will alter the original `x` from the calling function.

Using reference parameters is particularly useful when you need to return more than one value from a function or when the function's computations need to adjust the values of the variables passed to it. In our example, the updates on `v` directly overwrite the value of `x`, reflecting the changes outside the function after every iteration of the for-loop. This way, the program's flow can use and adapt to the new values of `x` seamlessly.