Assume 16-bit signed int: what hexadecimal is printed for this arithmetic right shift? #include<stdio.h> int main() { printf("%x ", -1 >> 1); return 0; }

Introduction / Context:
This question probes your understanding of right-shifting negative signed integers on a 16-bit implementation, and of how printf prints hexadecimal using %x. The key idea is arithmetic right shift on two’s-complement values.

Given Data / Assumptions:

• Assume int is 16 bits and two’s complement.
• -1 in two’s complement is 0xFFFF.
• Most compilers implement right shift of signed negative values as an arithmetic shift (propagating the sign bit).

Concept / Approach:
An arithmetic right shift fills in with the sign bit (1 for negative numbers). Shifting 0xFFFF right by any number of positions preserves all ones. Therefore, -1 >> 1 remains 0xFFFF when interpreted as a 16-bit signed arithmetic shift. Printing with %x emits lowercase hexadecimal without leading 0x.

Step-by-Step Solution:
Represent -1 → 0xFFFF.Perform arithmetic shift right by 1 → 0xFFFF (ones shift in).printf("%x", 0xFFFF) → prints ffff.

Verification / Alternative check:
Many compilers document that right shift of negative signed integers is implementation-defined, but on mainstream targets it is arithmetic. If it were a logical shift, the result would be 0x7FFF; the question’s 16-bit assumption targets the common arithmetic behavior.

Why Other Options Are Wrong:
0fff and fff0 do not correspond to a one-bit arithmetic right shift of all ones. 0000 is impossible for -1 >> 1 under arithmetic shifting. 7fff would be a logical right shift, not typical for signed.

Common Pitfalls:
Confusing logical vs arithmetic shifts; forgetting integer size assumptions; overlooking that %x prints unsigned representation of the resulting bit pattern.

Final Answer:
ffff