C++ reference semantics (post-increment in stream): what exact output appears and why?\n\n#include<iostream.h>\nint main()\n{\n int x = 10;\n int &y = x;\n x++;\n cout << x << " " << y++;\n return 0;\n}

Introduction / Context:
This problem ensures you understand references and post-increment when used inside a streaming expression. Because y is a reference to x, y++ modifies x after yielding its current value.

Given Data / Assumptions:

• Start: x=10.
• int &y = x; so y aliases x.
• Then x++; makes x=11.

Concept / Approach:
Post-increment (y++) yields the old value of y (which equals the current x) and then increments the object. Streaming prints values in order, using the produced value of each subexpression.

Step-by-Step Solution:
1) After x++, x becomes 11. 2) cout << x prints 11 first. 3) y++ yields 11 (the current value), then increments the same underlying x to 12. 4) The output tokens are thus 11 and 11; the final state is x=12.

Verification / Alternative check:
Replace y with x; x++ in the stream shows identical printed result but changes the final state after printing.

Why Other Options Are Wrong:
Options showing 12 as the first number ignore that increment happened before printing; options showing 12 as the second number forget that post-increment prints the old value.

Common Pitfalls:
Confusing pre- vs post-increment result value; forgetting that y and x are the same object.

Final Answer:
11 11