In C, a union shares storage among its members. For the union below, values are assigned sequentially to a and b. What will be printed when v.a is output? #include<stdio.h> int main() { union var { int a, b; }; union var v; v.a = 10; v.b = 20; printf("%d\n", v.a); return 0; }

Introduction / Context:
Unions in C overlay their members in the same memory location, so writing to one member overwrites the storage seen by the others. This question tests that understanding.

Given Data / Assumptions:

• union var { int a, b; } has members sharing storage.
• Assignments are v.a = 10; then v.b = 20.
• Print v.a afterwards.

Concept / Approach:
The last write into the union’s shared storage sets the bits seen by all members. After writing v.b = 20, both a and b see the same bit pattern, which corresponds to 20 when interpreted as int.

Step-by-Step Solution:

Verification / Alternative check:
Printing v.b immediately after would also show 20, confirming the shared memory behavior of unions.

Why Other Options Are Wrong:

• 10: Ignores the later overwrite by v.b = 20.
• 30: No arithmetic addition occurs; union does not combine values.
• 0: No zeroing of memory takes place here.

Common Pitfalls:
Assuming unions keep independent values like structs; they do not. Only the most recent write is represented in shared storage.

Final Answer:
20