C Macro & Preprocessor

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Question 1

#include <stdio.h>
#define PRINT(i, limit) do 
                        { 
                            if (i++ < limit) 
                            { 
                                printf("GeeksQuizn"); 
                                continue; 
                            } 
                        }while(1)

int main()
{
    PRINT(0, 3);
    return 0;
}

How many times GeeksQuiz is printed in the above program?

A	1
B	3
C	4
D	Compile-time error

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Question 1 Explanation:

The PRINT macro gets expanded at the pre-processor time i.e. before the compilation time. After the macro expansion, the if expression becomes: if (0++ < 3). Since 0 is a constant figure and represents only r-value, applying increment operator gives compile-time error: lvalue required. lvalue means a memory location with some address.

Question 2

#include <stdio.h>
#if X == 3
    #define Y 3
#else
    #define Y 5
#endif

int main()
{
    printf("%d", Y);
    return 0;
}

What is the output of the above program?

A	3
B	5
C	3 or 5 depending on value of X
D	Compile time error

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Question 2 Explanation:

In the first look, the output seems to be compile-time error because macro X has not been defined. In C, if a macro is not defined, the pre-processor assigns 0 to it by default. Hence, the control goes to the conditional else part and 5 is printed. See the next question for better understanding.

Question 3

What is the output of following program?

#include <stdio.h>
#define macro(n, a, i, m) m##a##i##n
#define MAIN macro(n, a, i, m)

int MAIN()
{
    printf("GeeksQuiz");
    return 0;
}

A	Compiler Error
B	GeeksQuiz
C	MAIN
D	main

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Question 3 Explanation:

The program has a preprocessor that replaces "MAIN" with "macro(n, a, i, m)". The line "macro(n, a, i, m)" is again replaced by main. The key thing to note is token pasting operator ## which concatenates parameters to macro.

Question 4

#include <stdio.h>
#define X 3
#if !X
    printf("Geeks");
#else
    printf("Quiz");
 
#endif
int main()
{
        return 0;
}

A	Geeks
B	Quiz
C	Compiler Error
D	Runtime Error

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Question 4 Explanation:

A program is converted to executable using following steps 1) Preprocessing 2) C code to object code conversion 3) Linking The first step processes macros. So the code is converted to following after the preprocessing step.

printf("Quiz");
int main()
{
        return 0;
}

The above code produces error because printf() is called outside main. The following program works fine and prints "Quiz"

#include 
#define X 3

int main()
{
#if !X
    printf("Geeks");
#else
    printf("Quiz");

#endif
    return 0;
}

Question 5

#include <stdio.h>
#define ISEQUAL(X, Y) X == Y
int main()
{
    #if ISEQUAL(X, 0)
        printf("Geeks");
    #else
        printf("Quiz");
    #endif
    return 0;
}

Output of the above program?

A	Geeks
B	Quiz
C	Any of Geeks or Quiz
D	Compile time error

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Question 5 Explanation:

The conditional macro #if ISEQUAL(X, 0) is expanded to #if X == 0. After the pre-processing is over, all the undefined macros are initialized with default value 0. Since macro X has not been defined, it is initialized with 0. So, Geeks is printed.

Question 6

#include <stdio.h>
#define square(x) x*x
int main()
{
  int x;
  x = 36/square(6);
  printf("%d", x);
  return 0;
}

A	1
B	36
C	0
D	Compiler Error

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Question 6 Explanation:

Preprocessor replaces square(6) by 6*6 and the expression becomes x = 36/6*6 and value of x is calculated as 36. Note that the macro will also fail for expressions "x = square(6-2)" If we want correct behavior from macro square(x), we should declare the macro as

#define square(x) ((x)*(x))

Question 7

Output?

# include <stdio.h>
# define scanf  "%s Geeks Quiz "
int main()
{
   printf(scanf, scanf);
   return 0;
}

A	Compiler Error
B	%s Geeks Quiz
C	Geeks Quiz
D	%s Geeks Quiz Geeks Quiz

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Question 7 Explanation:

After pre-processing phase of compilation, printf statement will become. printf("%s Geeks Quiz ", "%s Geeks Quiz "); Now you can easily guess why output is "%s Geeks Quiz Geeks Quiz".

Question 8

#include <stdio.h>
#define a 10
int main()
{
  printf("%d ",a);

  #define a 50

  printf("%d ",a);
  return 0;
}

A	Compiler Error
B	10 50
C	50 50
D	10 10

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Question 8 Explanation:

Preprocessor doesn't give any error if we redefine a preprocessor directive. It may give warning though. Preprocessor takes the most recent value before use of and put it in place of a.

Question 9

Output?

#include<stdio.h> 
#define f(g,g2) g##g2 
int main() 
{ 
   int var12 = 100; 
   printf("%d", f(var,12)); 
   return 0; 
}

A	100
B	Compiler Error
C	0
D	1

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Question 9 Explanation:

The operator ## is called “Token-Pasting” or “Merge” Operator. It merges two tokens into one token. So, after preprocessing, the main function becomes as follows, and prints 100.

int main() 
{ 
   int var12 = 100; 
   printf("%d", var12); 
   return 0; 
}

Question 10

Which file is generated after pre-processing of a C program?

A	.p
B	.i
C	.o
D	.m

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Question 10 Explanation:

After the pre-processing of a C program, a .i file is generated which is passed to the compiler for compilation.

There are 21 questions to complete.

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