Enumerations in C: determine the printed integral values for three students’ statuses. #include int main() { enum status { pass, fail, absent }; enum status stud1, stud2, stud3; stud1 = pass; stud2 = absent; stud3 = fail; printf("%d %d %d ", stud1, stud2, stud3); return 0; }

Introduction / Context: This checks understanding of default numbering of enum constants in C and how assigning them to variables results in specific integer outputs with printf.

Given Data / Assumptions:

• enum status defines pass, fail, absent in that order with no explicit values.
• Default enumeration starts at 0 and increments by 1.
• Assignments: stud1 = pass, stud2 = absent, stud3 = fail.

Concept / Approach: With no explicit assignments, pass = 0, fail = 1, absent = 2. The program prints the underlying integer values of the enumerators assigned to stud1, stud2, and stud3.

Step-by-Step Solution:

Verification / Alternative check: Adding explicit values (e.g., pass = 10) would shift subsequent values. With defaults, the simple 0/1/2 sequence applies.

Why Other Options Are Wrong:

• 0 1 2: Puts absent as 1; incorrect ordering for the given assignments.
• 1 2 3: Assumes start at 1 and increments; not how C enumerations default.
• 1 3 2: Not consistent with any default mapping.

Common Pitfalls: Mixing up the order of printing with the declaration order, or assuming enums start at 1 by default.

Final Answer: 0 2 1