C++ (const member function) — does a const-qualified Show() print the initialized member value?\n\n#include<iostream.h>\nclass Tab {\n int x;\npublic:\n Tab();\n void Show() const;\n ~Tab() {}\n};\nTab::Tab() { x = 5; }\nvoid Tab::Show() const { cout << x; }\nint main() {\n Tab objB; objB.Show();\n return 0;\n}\n\nPredict the output.

Introduction / Context:
This checks whether a const-qualified member function can read and print a data member that was assigned in the constructor. It reinforces the difference between reading (allowed in const methods) and writing (disallowed) to members.

Given Data / Assumptions:

• Constructor sets x = 5.
• Show() is declared const and prints x.
• Destructor is trivial and defined inline.

Concept / Approach:
A const member function promises not to modify any non-mutable data members. Reading x and sending it to cout is legal. Since x is assigned in the constructor, its value is well-defined when Show() executes.

Step-by-Step Solution:

Verification / Alternative check:
Marking Show() non-const would not change behavior here. Attempting to modify x inside Show() would fail to compile due to const qualification.

Why Other Options Are Wrong:

• Garbage value: x is initialized deterministically.
• Compile/runtime error: No rule is violated.

Common Pitfalls:
Assuming const methods can’t even read members or forgetting to initialize members before use.

Final Answer:
The program will print the output 5.