Q.Find and write the output of the following C++ program code: Note : Assume all required header files are already included in the program. void Convert(float &X, int Y=2) { X=X/Y; Y=X+Y; cout<<X<<"*"<<Y<<endl; } void main() { float M=15, N=5; Convert(M,N); Convert(N); Convert(M); }

The key idea involves pass by reference, default function arguments, and implicit type conversions in C++.

Here is the step-by-step execution:

1. Convert(M, N): M (15.0) is passed by reference to X, and N (5) is passed by value to Y.

   • X = X / Y; becomes M = 15.0 / 5; so M becomes 3.0.
   • Y = X + Y; becomes Y = 3.0 + 5; so Y becomes 8 (integer truncation from 8.0).
   • The program prints 3.0*8. M is now 3.0, N remains 5.0.

2. Convert(N): N (5.0) is passed by reference to X. Y uses its default value 2.

   • X = X / Y; becomes N = 5.0 / 2; so N becomes 2.5.
   • Y = X + Y; becomes Y = 2.5 + 2; so Y becomes 4 (integer truncation from 4.5).
   • The program prints 2.5*4. N is now 2.5. …

The core concepts at play here are call-by-reference for the first parameter X and default arguments for the second parameter Y. Understanding how these work is key to tracing the program's execution.

When a parameter is passed by reference (like float &X), any changes made to X inside the Convert function directly modify the original variable in the main function that was passed to it. This is different from call-by-value, where a copy is made, and the original variable remains unchanged.

The int Y=2 in the function signature means that if the Convert function is called with only one argument, the compiler will automatically use 2 as the value for Y. If two arguments are provided, the second argument will override this default value.

Let's trace the program's execution step-by-step:

1. Initial State in main:

   • float M = 15;
   • float N = 5;

   At this point, M holds 15.0 and N holds 5.0.

2. First Call: Convert(M, N);

   • The Convert function is called.
   • X (a reference) is bound to M. So, X now refers to M's memory location, and its initial value is 15.0.
   • Y (a value parameter) receives the value of N. So, Y is 5.
   • Inside Convert:
     • X = X/Y;
       • This becomes M = 15.0 / 5;
       • M (and X) is updated to 3.0.
     • Y = X+Y;
       • This becomes Y = 3.0 + 5;
       • Y (the local parameter) is updated to 8.
     • cout<<X<<"*"<<Y<<endl;
       • This prints the current value of X (which is 3.0) followed by * and then the current value of Y (which is 8).
       • Output: 3*8

   After this call, M is 3.0, and N remains 5.0. The local Y inside Convert is destroyed.

3. Second Call: Convert(N);

   • The Convert function is called.
   • X (a reference) is bound to N. So, X now refers to N's memory location, and its initial value is 5.0.
   • Since only one argument is provided, Y takes its default value of 2.
   • Inside Convert:
     • X = X/Y;
       • This becomes N = 5.0 / 2;
       • N (and X) is updated to 2.5.
     • Y = X+Y;
       • This becomes Y = 2.5 + 2;
       • Y (the local parameter) is updated to 4.
     • cout<<X<<"*"<<Y<<endl;
       • This prints the current value of X (which is 2.5) followed by * and then the current value of Y (which is 4).
       • Output: 2.5*4

   After this call, M is 3.0, and N is 2.5. The local Y inside Convert is destroyed. …