C (16-bit unsigned int assumed): what does this program print? #include<stdio.h> int main() { unsigned int i = 65536; /* Assume 2 byte integer */ while (i != 0) printf("%d", ++i); printf(" "); return 0; }

Introduction / Context:This question targets integer width assumptions and initialization effects. Assigning a value outside the range of a 16-bit unsigned type causes wraparound, which directly determines whether the loop body executes at all.

Given Data / Assumptions:

• unsigned int is assumed to be 16 bits (0..65535).
• 65536 is exactly 2^16, so it wraps to 0 on assignment in a 16-bit unsigned int.
• The while loop executes only if i != 0.

Concept / Approach:On a 16-bit model, unsigned int i = 65536; stores 0 because 65536 mod 65536 = 0. Therefore the loop condition (i != 0) is false at the first test, and the loop body is skipped entirely. Only the trailing newline is printed, which most judges interpret as “no visible output.”

Step-by-Step Solution:

Verification / Alternative check:Change the initializer to 65535 and observe the loop execute; or print the value of i immediately after initialization to confirm it is 0.

Why Other Options Are Wrong:

Common Pitfalls:Forgetting the target width of unsigned int on 16-bit assumptions; assuming 32-bit widths everywhere.

Final Answer:No output.