On a 16-bit platform, what is sizeof for this enumeration variable?\n\n#include <stdio.h>\n\nint main()\n{\n enum value { VAL1 = 0, VAL2, VAL3, VAL4, VAL5 } var;\n printf("%d\n", (int)sizeof(var));\n return 0;\n}\n

Introduction / Context:
This problem checks knowledge of how enumeration types are stored in memory on older 16-bit C implementations and what sizeof yields for an enum object.

Given Data / Assumptions:

• Target platform is 16-bit.
• enum value has five named constants.
• We are printing sizeof(var).
• Assume conventional 16-bit compilers where int is 2 bytes and enum has the same size as int.

Concept / Approach:
In C, the size of an enum type is implementation-defined, but historically on 16-bit compilers (e.g., Turbo C), enum typically has the size of int. On such systems, int is 2 bytes, so sizeof(enum variable) is 2.

Step-by-Step Solution:

Verification / Alternative check:
Modern 32-bit or 64-bit compilers often make enum the size of int (commonly 4 bytes). The question explicitly fixes a 16-bit context to remove this ambiguity.

Why Other Options Are Wrong:

• 1: Too small; not typical for an enum mapped to int.
• 4: Common on 32-bit/64-bit, but not the stated 16-bit case.
• 10: Not meaningful for sizeof an enum variable here.

Common Pitfalls:
Assuming sizeof(enum) is always 4. It depends on the compiler and target model; always consider platform details given in the stem.

Final Answer:
2