C++ reference pre-decrement through an alias: what is printed?\n\n#include<iostream.h>\nint main()\n{\n int x = 80;\n int &y = x;\n x++;\n cout << x << " " << --y;\n return 0;\n}

Introduction / Context:
This question verifies understanding of pre-decrement applied via a reference. Because y aliases x, changing y changes x immediately.

Given Data / Assumptions:

• Initial x=80, then x++ gives x=81.
• y refers to the same object as x.

Concept / Approach:
Pre-decrement (--y) decrements first and yields the decremented value. Since y aliases x, --y reduces x to 80 and yields 80.

Step-by-Step Solution:
1) After x++, x is 81. 2) First printed value: x is 81. 3) --y decrements the same object to 80 and yields 80; second printed value is 80. 4) Final state: x=80.

Verification / Alternative check:
If --x were used instead, the second printed value would still be 80, showing the alias equivalence.

Why Other Options Are Wrong:
80 80 ignores the prior increment; 81 81 misapplies pre-decrement; compile error is inapplicable.

Common Pitfalls:
Forgetting that y is an alias for x and that pre-decrement changes the value before yielding it.

Final Answer:
81 80