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 Alter(char *S1, char *S2) { char T; T=S1; S1=S2; S2=T; cout<<S1<<"&"<<S2<<end1; } void main() { char X[]="First", Y[]="Second"; Alter(X,Y); cout<<X<<""<<Y<<end1; }

Concept: Pointer swapping vs. array content — swapping pointers inside a function only changes the local copies, not the original arrays.

Reasoning:

1. Alter(X,Y) passes the base addresses of arrays X and Y into the pointers S1 and S2.
2. Inside Alter, the pointers S1 and S2 are swapped using T, so S1 now points to "Second" and S2 points to "First". The cout prints Second&First. …

The trap here is subtle but classic: when you pass an array to a function in C++, what actually gets passed is a pointer to its first element. The function receives a copy of that pointer. Swapping the copies inside the function has no effect on the original arrays in main.

Let's trace it carefully.

1. What gets passed to Alter?

   X and Y are character arrays. When used as arguments, they "decay" to pointers: X becomes a pointer to the first 'F', and Y becomes a pointer to the first 'S'. The function receives copies of these pointer values — call them S1 and S2. So initially, S1 points to "First" and S2 points to "Second".

2. Inside Alter, the swap happens on the copies.

   A temporary pointer T is used to swap S1 and S2. After the swap:

   • S1 now points to "Second"
   • S2 now points to "First"

   Then the function prints S1 << "&" << S2, which outputs Second&First.

3. Back in main, nothing has changed.

   The original arrays X and Y still hold their original contents. The pointers that were swapped were just local copies inside Alter. So cout << X << "*" << Y prints First*Second. …