Conditional compilation with #ifdef – does this program compile and what is printed?\n\n#include <stdio.h>\n\nint main()\n{\n#ifdef NOTE\n int a; a = 10;\n#else\n int a; a = 20;\n#endif\n printf("%d\n", a);\n return 0;\n}

Introduction / Context:
Conditional compilation via #ifdef allows different code to be compiled depending on whether a macro is defined. This example tests whether duplicate declarations occur and what value is printed when the macro is not defined.

Given Data / Assumptions:

• NOTE is not defined anywhere.
• Both branches declare int a; and assign distinct values.
• Only one branch is compiled and the other is discarded.

Concept / Approach:
The preprocessor selects exactly one branch: since NOTE is undefined, the #else path is included. In the final translation unit, there is a single definition and assignment of a. There is no duplication because the excluded branch is not part of the compiled code.

Step-by-Step Solution:
Preprocess: retain the #else block only.Resulting code defines a and sets a = 20.Print statement outputs 20.

Verification / Alternative check:
Add #define NOTE before #ifdef and observe that the output becomes 10. Using a compiler flag that defines a macro (e.g., -DNOTE) also flips the branch.

Why Other Options Are Wrong:
Duplicate definition — incorrect because only one branch is compiled. Require prior #define — not required; it simply chooses the #else branch. Prints 10 — would occur only if NOTE were defined.

Common Pitfalls:
Declaring variables in both branches but using them outside without consistent types; forgetting that preprocessor removes the inactive branch entirely.

Final Answer:
Yes — it compiles and prints 20