Multiple choice

main() {
extern i; printf("%d",i); { int i=20; printf("%d",i); } }

  1. 20 20

  2. Garbage Value 20

  3. 0 20

  4. Linker Error

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The variable i is declared as extern but never defined anywhere in the program. This causes a linker error because the linker cannot find the definition of i to satisfy the external declaration. The inner block's int i=20 is a separate local variable and doesn't resolve the extern.