C++ reference aliasing and pre-decrement: what exact numbers are printed?\n\n#include <iostream.h>\nint main() {\n int x = 10;\n int &y = x;\n x = 25;\n y = 50; // x now 50 as well\n cout << x << " " << --y; // print x, then pre-decrement y (and thus x)\n return 0;\n}\n

Introduction / Context:
The item checks precise evaluation order when printing a referenced variable and then applying a pre-decrement to that same aliased object. Understanding that y is an alias of x is critical here.

Given Data / Assumptions:

• y is a reference bound to x.
• After assignments, x == 50 and y refers to x.
• cout << x occurs before evaluating --y in the chained insertion expressions.

Concept / Approach:
In the chained output, operators are sequenced left-to-right. First, the current value of x (50) is sent to the stream. Next, --y is evaluated, which decrements the aliased object from 50 to 49, and that value (49) is inserted. Although x becomes 49 due to the aliasing, x was already printed as 50 prior to the decrement.

Step-by-Step Solution:

Verification / Alternative check:
Replace --y with y-- and observe a different result (50 50) because y-- prints the old value before decrement; here we use pre-decrement.

Why Other Options Are Wrong:

• 25 49: ignores updates to 50.
• 50 50: would require post-decrement.
• 49 49: would require printing after the decrement for both positions.
• Compile time error: code is valid under classic headers.

Common Pitfalls:
Assuming both outputs reflect the decremented value; confusing pre- and post-decrement semantics.

Final Answer:
The program will print the output 50 49.