PREPROCESSOR DIRECTIVES : BEAUTY OF C LANGUAGE

Monday, 30 July 2012 | Unknown at 09:14 |

(1) What will be output of following code?

#include<stdio.h>

#define max 10

int main(){

    int i;

    i=++max;

    printf("%d",i);

    return 0;

}

(2) What will be output of following code?

#include<stdio.h>

#define max 10+2

int main(){

    int i;

    i=max*max;

    printf("%d",i);

    return 0;

}

(3) What will be output of following code?

#include<stdio.h>

#define A 4-2

#define B 3-1

int main(){

     int ratio=A/B;

     printf("%d ",ratio);

     return 0;

}

(4) What will be output of following code?

#include<stdio.h>

#define MAN(x,y) (x)>(y)?(x):(y)

int main(){

       int i=10,j=9,k=0;

       k=MAN(i++,++j);

       printf("%d %d %d",i,j,k);

       return 0;

}

(5) What will be output of following code?

#include<stdio.h>

#define START main() {

#define PRINT printf("*******");

#define END }

START

PRINT

END

(6) What will be output of following code?

#define CUBE(x) (x*x*x)

#define M 5

#define N M+1

#define PRINT printf("Daarks Solutions & Services");

int main(){

      int volume =CUBE(3+2);

      printf("%d %d ",volume,N);

      PRINT

      return 0;

}

Solution section 

(1)output: compiler error.

Explanation:

Here max is preprocessor macro symbol which
 process first before

 the actual compilation. First preprocessor 
replace the symbol to its 

value in entire the program before the compilation. 

So in this program max will be replaced by 10 before
 compilation. 

Thus program will be converted like this:

int main(){

     int i;

     i=++10;

     printf("%d",i);

     return 0;

}

To know how we can see intermediate file click here

In this program we are trying to increment a constant 
symbol.

Meaning of ++10 is:

10=10+1

or 10=11

Which is error because we cannot assign constant value to another

 constant value .Hence compiler will give error.

(2)

Output: 32

Explanation:

Here max is preprocessor macro symbol which process
 first before

 the actual compilation start. Preprocessor replace
 the symbol to 

its value in entire the program before the compilation. 

So in this program max will be replaced by 10+2 before
 compilation.

 Thus program will be converted as:

int main(){

    int i;

    i=10+2*10+2;

    printf("%d",i);

    return 0;

}

now i=10+2*10+2

i=10+20+2

i=32

(3)

Output: 3

Explanation:

A and B are preprocessor macro symbol which process first 
before the actual 

compilation start. Preprocessor replace the symbol to its value in 

entire the program before the compilation.

 So in this program A and B will be replaced by 4-2 and 3-1 respectively 

 before compilation. Thus program will be converted as: 

int main(){

    int ratio=4-2/3-1;

    printf("%d ",ratio);

    return 0;

}

Here ratio=4-2/3-1

ratio=4-0-1

ratio=3

(4)

Output: 11 11 11

Explanation:

Preprocessor’s macro which process first before the actual compilation.

 Thus program will be converted as: 

int main(){

    int i=10,j=9,k=0;

    k=(i++)>(++j)?(i++):(++j);

    printf("%d %d %d",i,j,k);

    return 0;

}

now k=(i++)>(++j)?(i++):(++j);

first it will check the condition

(i++)>(++j)

i++ i.e. when postfix is used with variable in 
expression then 

expression is evaluated first with original value 
then variable

 is incremented

Or 10>10

This condition is false.

Now i = 10+1 = 11

There is rule, only false part will execute after? i.e. ++j, i++ will be not execute.

So after ++j

j=10+1=11;

And k will assign value of j .so k=11; 

(5)

Output: *******

Explanation:

This program will be converted as: 

main(){

    printf("*******");

}

(6)

Output: 17 6

Explanation: This program will be converted as:

int main(){

    int volume =(3+2*3+2*3+2);

    printf("%d %d ",volume,5+1);

    printf("Daarks Solutions & Services");

    return 0;

}