String literals and macros: what does this program print? #include #define MESS junk int main() { printf("MESS "); return 0; } Does the macro affect the content of the string literal?

Introduction / Context:Preprocessor macros perform textual substitution before compilation, but they do not modify the contents of string literals. This question verifies that understanding.

Given Data / Assumptions:

• Macro definition: #define MESS junk.
• printf is called with a string literal "MESS".
• No macro is used outside the quotes in the call to printf.

Concept / Approach:During preprocessing, identifiers inside string literals are not expanded. Only tokens outside string literals are candidates for macro substitution.

Step-by-Step Solution:Preprocessor sees printf("MESS"); unchanged.At runtime, printf prints the literal text MESS followed by a newline.

Verification / Alternative check:If the code were printf("%s", MESS); without quotes around MESS, then the token MESS would expand to junk and compilation would likely fail (undefined identifier) unless junk were defined as a string.

Why Other Options Are Wrong:"junk": would require macro expansion inside the string literal, which does not occur.Error / nothing: the code is valid and prints text.

Common Pitfalls:Expecting the preprocessor to parse inside quotes; forgetting that stringization requires the # operator within a macro definition.

Final Answer:MESS