C 16-bit integer constants: Which of the following is not a valid 16-bit integer constant value in C (assuming a 16-bit int range of −32768 to +32767)?

Question

Solve this multiple-choice question and choose the correct option.

Curioustab · Accepted Answer

Introduction / Context:This tests knowledge of integer literal ranges for a 16-bit int. While modern systems typically use 32-bit int, many exam questions still assume a 16-bit range for portability and historical reasons. Given Data / Assumptions: Assumed 16-bit signed int range: −32768 to +32767. We examine each literal to see whether it lies within the permitted range. Concept / Approach:Any literal outside the stated range is invalid for assignment to a 16-bit int without causing overflow or requiring a larger integer type. The syntax of the constants is otherwise standard (leading + or − sign allowed). Step-by-Step Solution: 321, −162, +1, and 0 are clearly within −32768…+32767.−65321 is much less than −32768; it is out of range for a 16-bit int.Hence, −65321 is not a valid 16-bit integer constant under the assumption. Verification / Alternative check:Consider storage width: 16-bit two’s complement cannot encode magnitude 65321; wrap-around would occur if forced, which is undefined/implementation-specific in literal typing context. Why Other Options Are Wrong: All other values fall within the representable range and thus are valid. Common Pitfalls: Forgetting that the negative bound (−32768) has magnitude one larger than the positive bound (+32767) in two’s complement. Final Answer: -65321

C 16-bit integer constants: Which of the following is not a valid 16-bit integer constant value in C (assuming a 16-bit <code>int</code> range of −32768 to +32767)?

Answer

Explanation