L1.8 Preprocessor and function pointers · GATE 2027 Programming and Data Structures

ACompressed recap

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The preprocessor rewrites the source before compilation. #define replaces text: expand it completely, then evaluate with normal precedence. Bracket every parameter and the whole body, and never pass side effects. A function pointer int (*f)(int, int) stores a function’s address; call it as f(…) or (*f)(…); arrays of them and callbacks follow.

BMerged rule card

1.8 Preprocessor macros and their expansion traps; function pointers

Preprocessor runs before compiling: #include, #define, #ifdef.
#define is text substitution: expand fully, then apply precedence.
Bracket parameters and the whole body: #define SQ(x) ((x) * (x)).
Side-effect arguments may be evaluated several times (undefined for ++).
int (*f)(int, int) = add; f(2, 3) or (*f)(2, 3).
Function pointer arrays and callbacks; int *f(int) is different.

CInterleaved GATE-style questions

Q1NAT1 markMacro

With #define DOUBLE(x) x + x, what is 3 * DOUBLE(2)?

Numerical answer: ________

Q2NAT2 marksMacro

With #define N 3 and #define F N + 1, what is F * F?

Numerical answer: ________

Q3MCQ2 marksOutput

What does this print?

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define N 4
#define AREA N * N + 1

int main(void)
{
    int x = 3, y = 7;
    printf("%d %d %d\n", MAX(x, y), 2 * AREA, MAX(x + 1, y - 4));
    return 0;
}

(A) 7 33 3
(B) 7 33 4
(C) 7 34 4
(D) 3 33 4

Q4MCQ1 markFunction pointer call

If int (*f)(int) = g;, which calls g with 5?

(A) *f(5)
(B) f(5)
(C) f[5]
(D) &f(5)

Q5MCQ1 markBrackets

Which definition makes 1000 / SQ(10) equal to 10?

(A) #define SQ(x) (x) * (x)
(B) #define SQ(x) (x * x) and also (x) unused
(C) #define SQ(x) x * x
(D) #define SQ(x) ((x) * (x))

Q6MSQ2 marksFacts

Which are true?

(A) A function-like macro is type-checked like a function.
(B) Function pointers can be stored in arrays.
(C) #define creates no variable in memory.
(D) A macro can be redefined after #undef.

Q7MCQ2 marksOutput

What does this print?

#define PROD(a, b) a * b
#define DOUBLE(x) x + x

int main(void)
{
    int k = 2;
    printf("%d %d\n", PROD(k + 1, k + 2), 3 * DOUBLE(k));
    return 0;
}

(A) 6 12
(B) 12 12
(C) 6 8
(D) 7 8

Q8NAT1 markCallback

With apply(mul, 6, 7) from Lesson 1.8, what is returned?

Numerical answer: ________

DAnswer Vault