What happens when #if 0...#endif wraps a block of code?
Answer: B
#if 0...#endif is used to disable code during preprocessing. The enclosed code is skipped entirely and doesn't appear in the compiled output. This is more efficient than C-style comments for large blocks and maintains nested comment compatibility.
Q.962Hard
Which of the following will cause infinite recursion when used?
#define RECURSE() RECURSE()
Answer: C
The C preprocessor does not detect or prevent recursive macro expansion in its definition. RECURSE() will expand to RECURSE() infinitely, causing the preprocessor to hang or run out of memory. Modern compilers have safeguards, but theoretically this creates infinite expansion.
Q.963Easy
What is the purpose of defined() operator in preprocessor conditionals?
Answer: A
defined(MACRO) is used in #if and #elif directives to test if a macro is defined. Example: #if defined(DEBUG) ... #endif. It's equivalent to #ifdef but can be used in expressions with logical operators like && and ||.
Q.964Medium
Consider: #define DOUBLE(x) (2*(x))
int main() { int arr[DOUBLE(5)]; ... }
What is the size of the array?
Answer: B
The macro DOUBLE(5) expands to (2*(5)) = 10 at preprocessing time. Array declarations require compile-time constant expressions, and macro-expanded constants qualify. The array has 10 elements. This works because the macro creates a constant expression that the compiler can evaluate.
Q.965Easy
What is the primary purpose of the C preprocessor?
Answer: A
The preprocessor is a text processing tool that processes source code before actual compilation, handling directives like #define, #include, and macros.
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Q.966Easy
Which directive is used to conditionally compile code based on a logical condition?
Answer: B
#if evaluates constant expressions for conditional compilation. #ifdef checks if a macro is defined, #define creates macros, and #pragma provides compiler-specific directives.
Q.967Medium
What will be the preprocessed output of the following code?
#define SQUARE(x) x*x
int result = SQUARE(5+3);
Answer: C
Without parentheses, SQUARE(5+3) expands to 5+3*5+3, which evaluates to 5+15+3=23 due to operator precedence, not (5+3)*(5+3)=64. This demonstrates why macro parameters need parentheses.
Q.968Medium
What does the __VA_ARGS__ preprocessor feature allow in variadic macros?
Answer: A
__VA_ARGS__ allows macros to accept a variable number of arguments. Example: #define PRINT(...) printf(__VA_ARGS__) enables flexible argument passing.
Q.969Medium
What is the output of the following code?
#define STR(x) #x
printf(STR(Hello World));
Answer: B
The # operator (stringification) converts the macro argument into a string literal. STR(Hello World) becomes "Hello World", which printf prints with quotes.
Q.970Medium
Consider this preprocessor code:
#define CONCAT(a,b) a##b
int CONCAT(var,1) = 10;
What is the actual variable name created?
Answer: B
The ## operator (token pasting) concatenates tokens directly. CONCAT(var,1) produces var1 as the variable name, not a##b.
Q.971Hard
What will happen if a macro is defined multiple times with different definitions in the same compilation unit?
#define SIZE 10
#define SIZE 20
Answer: A
Redefining a macro with a different value in the same compilation unit causes a compilation error. To redefine, you must #undef first.
Q.972Medium
Consider the following preprocessor directives:
#define MAX 100
#define MIN 50
#undef MAX
#define MAX 200
int main() {
printf("%d", MAX);
return 0;
}
What will be the output of this program?
Answer: A
The #undef directive undefines the previously defined macro MAX (which was 100). Then MAX is redefined as 200. So printf will output 200. The #undef allows redefining macros without compilation errors.
