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

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