C++ references and side effects inside a while loop: trace the output step by step.\n\n#include <iostream.h>\nint main() {\n int x = 0;\n int &y = x; y = 5;\n while (x <= 5) {\n cout << y++ << " ";\n x++;\n }\n cout << x;\n return 0;\n}\n

Introduction / Context:
This program demonstrates how a reference (y) bound to x mirrors and mutates x. It also illustrates the interaction between y++ (which increments x) and a separate x++ within the same loop body, changing the loop's termination sooner than expected.

Given Data / Assumptions:

• y is a reference to x.
• y = 5 sets x to 5.
• Loop condition: while (x <= 5).
• Body: prints y++ then executes x++.

Concept / Approach:
Since y aliases x, the expression y++ increments x. The loop body therefore increments x twice per iteration: once via y++ and once via the explicit x++. This causes the loop to execute only once, printing the original value (5), and then exit with x == 7, which is printed after the loop.

Step-by-Step Solution:

Verification / Alternative check:
Removing the inner x++ would allow multiple prints: 5 6 7 8 9 10 and then 11 after loop; the extra x++ is the reason for the single-iteration behavior.

Why Other Options Are Wrong:

• Sequences showing many numbers assume only one increment per iteration.
• Compile error: code is valid.

Common Pitfalls:
Forgetting that y++ increments x; overlooking that two increments happen in one iteration and terminate the loop early.

Final Answer:
The program will print the output 5 7.