C++ globals and references: follow constructor prints and later mutations—what is the final console output? #include<iostream.h> int x, y; class CuriousTabTest { public: CuriousTabTest(int xx = 0, int yy = 0) { x = xx; y = yy; Display(); } void Display() { cout << x << " " << y << " "; } }; int main() { CuriousTabTest obj(10, 20); // prints once in ctor int &rx = x; int &ry = y; ry = x; // y = 10 rx = y; // x = 10 cout << rx--; // prints 10 then decrements x to 9 return 0; }

Introduction / Context:
This problem asks you to combine constructor-time printing, global variables, and references used later in main. The key is that the constructor prints immediately and later streaming of rx-- shows the value before decrement.

Given Data / Assumptions:

• Globals x and y.
• Constructor assigns x=10, y=20 and prints \"10 20 \".
• References rx and ry alias x and y.

Concept / Approach:
Post-decrement returns the current value then decrements the object. The intermediate assignments make both globals equal to 10 before the final print.

Step-by-Step Solution:
1) Constructor prints 10 20 . 2) ry = x; sets y=10. 3) rx = y; sets x=10. 4) cout << rx--; prints 10, then decrements x to 9. 5) Combined console output: 10 20 10 (spaces as shown).

Verification / Alternative check:
Change the last line to cout << --rx; to see 9 printed instead, matching option C's last number and highlighting the difference between pre and post operators.

Why Other Options Are Wrong:
0 0 10 contradicts constructor assignment; 10 20 9 confuses pre/post decrement; compile error is inapplicable; 20 10 10 swaps constructor values.

Common Pitfalls:
Overlooking that the constructor already printed; forgetting post-decrement prints the old value.

Final Answer:
10 20 10