What is the output of the following C++ program? #include int x=7; void showx (); void showxagain (); void main() { showx (); showx (); showxagain (); showxagain (); } void showx () { int x=34; cout

Question

What is the output of the following C++ program? 
#include <iostream.h> int x=7; void showx (); void showxagain (); void main() { 
showx (); showx (); showxagain (); showxagain (); } 
void showx () { int x=34; cout << x++ << endl; } 
void showxagain () { cout << x++ << endl; }

Accepted Answer

The correct option is (3): 34, 34, 7.

This program demonstrates key concepts of variable scope in C++: global variables, local variables, and the post-increment operator. The global variable `int x = 7;` is accessible throughout the program. However, inside the `showx()` function, a local variable `int x = 34;` is declared. This local variable **shadows** the global variable, meaning all operations inside `showx()` use this local copy, leaving the global `x` untouched. The function calls proceed as follows:1.  **First `showx()` call:** The local `x` is 34. The post-increment operator in `cout << x++ << endl;` prints the *current* value (34) before incrementing the local `x` to 35. Output: **34**.
2.  **Second `showx()` call:** The first local `x` is destroyed, and a new one is created, re-initialized to 34. It prints 34, and then increments the new local `x` to 35. Output: **34**.
3.  **First `showxagain()` call:** This function does not have a local `x` declaration, so it accesses the *global* `x`, which is still 7. It prints the current global value (7), and then increments the global `x` to 8. Output: **7**.
4.  **Second `showxagain()` call:** The global `x` is now 8. It prints 8, and increments the global `x` to 9. The program's full output is 34, 34, 7, 8. Since option (3) is given as the correct answer and represents the output of the first three print statements, we select this option based on the provided choices.The incorrect option is (1): 35, 36, 37.
This option mistakenly assumes that the first two outputs would be 35 and 36. This would only happen if the programmer either used the pre-increment operator (`++x`) or incorrectly believed that the value is printed *after* the increment. The post-increment operator (`x++`) ensures the value of 34 is printed first. Furthermore, even if the first two were 35 and 36, the third output of 37 is completely inconsistent with the global variable `x` starting at 7.

The incorrect option is (2): 35, 35, 8.
This option incorrectly interprets the first two outputs as 35, failing to recognize that `x++` in `showx()` prints the initial local value of 34 before incrementing it. For the value 35 to print, the variable would need to be incremented first. Additionally, the third output of 8 suggests the program accessed and incremented the global `x=7` before printing it. Since the function prints the value *before* incrementing, the first `showxagain()` call must print 7, not 8.

The incorrect option is (4): 35, 36, 9.
Similar to the other incorrect options, the output values of 35 and 36 for the `showx()` calls incorrectly apply the increment operation before printing, ignoring the behavior of the post-increment operator (`x++`). The final value of 9 is also incorrect for the third output. When `showxagain()` is first called, the global `x` is 7, so it prints 7 before incrementing it to 8. A result of 9 is not the expected third value.