Token pasting (##) in macros: what does this program print? #include<stdio.h> #define FUN(i, j) i##j int main() { int va1 = 10; int va12 = 20; printf("%d\n", FUN(va1, 2)); return 0; }

Introduction / Context:
Token pasting (##) concatenates tokens at preprocessing time to form a new token. This question checks whether you recognize that concatenating va1 with 2 yields the identifier va12, which is a different variable with its own value.

Given Data / Assumptions:

• Variables: va1 = 10, va12 = 20.
• Macro: #define FUN(i, j) i##j.
• Call: FUN(va1, 2).

Concept / Approach:
The macro concatenates the tokens va1 and 2 to form va12. Since va12 exists and is initialized to 20, the printf prints 20. If va12 did not exist, this would be a compile-time error, showing that token pasting occurs before semantic checks.

Step-by-Step Solution:

Verification / Alternative check:
Try FUN(va, 12) with int va12 defined, or change the numbers to see compile failures when the pasted identifier is undefined.

Why Other Options Are Wrong:

Common Pitfalls:
Confusing token pasting with runtime string operations; expecting arithmetic with 1 and 2 instead of identifier concatenation.

Final Answer:
20.