C macros with stringification: what does this program print? #include #define JOIN(s1, s2) printf("%s=%s %s=%s ", #s1, s1, #s2, s2) int main() { char *str1 = "India"; char *str2 = "CURIOUSTAB"; JOIN(str1, str2); return 0; }

Introduction / Context:This macro uses the C preprocessor’s stringification operator # to print both the parameter names and their string values. Understanding # helps in building debugging and logging helpers that display variable names with their contents.

Given Data / Assumptions:

• #define JOIN(s1, s2) printf("%s=%s %s=%s", #s1, s1, #s2, s2)
• str1 = "India", str2 = "CURIOUSTAB"
• Valid C compilation environment.

Concept / Approach:The stringification operator # converts the macro argument token sequence into a string literal. Hence #s1 becomes "str1" and #s2 becomes "str2". The other arguments (without #) pass the variable values, which are pointers to the respective strings. The final output is the variable names and their stored strings.

Step-by-Step Solution:

Verification / Alternative check:Change str1 to another variable name (e.g., country) and observe that the left-hand label changes accordingly.

Why Other Options Are Wrong:

Common Pitfalls:Assuming # stringifies the value instead of the token; swapping names and values.

Final Answer:str1=India str2=CURIOUSTAB.