In C, what output is produced by the following enumeration and assignments (default enum starting at 0)? #include<stdio.h> int main() { enum status { pass, fail, atkt }; enum status stud1, stud2, stud3; stud1 = pass; // 0 stud2 = atkt; // 2 stud3 = fail; // 1 printf("%d, %d, %d ", stud1, stud2, stud3); return 0; }

Introduction / Context:
C enumerations by default assign integer values starting at 0 and increment by 1 unless explicitly set. Understanding this default behavior helps in reading printed enum values and interfacing with APIs.

Given Data / Assumptions:

• enum status { pass, fail, atkt }; with no explicit initializers.
• Assignments: stud1 = pass; stud2 = atkt; stud3 = fail;
• Printing via printf("%d, %d, %d", ...)

Concept / Approach:
By default: pass = 0, fail = 1, atkt = 2. The program prints the integer values corresponding to the assigned enumerators.

Step-by-Step Solution:
Assign defaults: pass→0, fail→1, atkt→2.Map variables: stud1=0, stud2=2, stud3=1.Print order is stud1, stud2, stud3 → “0, 2, 1”.

Verification / Alternative check:
Adding explicit initializers (e.g., pass=10) would change the printed numbers accordingly; here no such initialization exists.

Why Other Options Are Wrong:
A/D/E list other permutations not matching the actual assignments. B starts at 1 which is not the default.

Common Pitfalls:
Assuming enums start at 1; forgetting the order of printing vs declaration order.

Final Answer:
0, 2, 1