C Language Placement Papers

1. What will be the output?

#include<stdio.h>
#define SWAP(a, b, c)(c t; t=a, a=b, b=t)
int main()
{
    int x=10, y=20;
    SWAP(x, y, int);
    printf("%d %d\n", x, y);
    return 0;
}
 
A.    It compiles
B.    Compiles with an warning
C.    Not compile
D.    Compiles and print nothing
    (NB:The code won't compile since declaration of t cannot occur within parenthesis. )

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2.In which stage the following code
#include<stdio.h>
gets replaced by the contents of the file stdio.h
A.     During editing    B.     During linking    C.     During execution    D.     During preprocessing
 (NB:The preprocessor replaces the line #include <stdio.h> with the system header file

of that name. More precisely, the entire text of the file 'stdio.h' replaces the  #include directive. ) 

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3.   What will be the output of the program?
#include<stdio.h>
#define MAN(x, y) ((x)>(y)) ? (x):(y);

int main()
{
    int i=10, j=5, k=0;
    k = MAN(++i, j++);
    printf("%d, %d, %d\n", i, j, k);
    return 0;
}

A.     12, 6, 12 ,   B.     11, 5, 11 ,  C.     11, 5, Garbage ,   D.     12, 6, Garbage

(NB:The macro MAN(x, y) ((x)>(y)) ? (x):(y); returns the biggest number of given two numbers.

Step 1: int i=10, j=5, k=0; The variable i, j, k are declared as an integer type and initialized to value 10, 5, 0 respectively.

Step 2: k = MAN(++i, j++); becomes,

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

=> k = ((11)>(5)) ? (12):(6);

=> k = 12

Step 3: printf("%d, %d, %d\n", i, j, k); It prints the variable i, j, k.

In the above macro step 2 the variable i value is increemented by 2 and variable j value is increemented by 1.

Hence the output of the program is 12, 6, 12)

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 4.What will be the output of the program?

#include<stdio.h>
#define PRINT(int) printf("int=%d, ", int);

int main()
{
    int x=2, y=3, z=4;  
    PRINT(x);
    PRINT(y);
    PRINT(z);
    return 0;
}

A.     int=2, int=3, int=4                          B.     int=2, int=2, int=2
C.     int=3, int=3, int=3                            D.     int=4, int=4, int=4
(NB:The macro PRINT(int) print("%d,", int); prints the given variable value in an integer format.

Step 1: int x=2, y=3, z=4; The variable x, y, z are declared as an integer type and initialized to 2, 3, 4 respectively.

Step 2: PRINT(x); becomes printf("int=%d,",x). Hence it prints 'int=2'.

Step 3: PRINT(y); becomes printf("int=%d,",y). Hence it prints 'int=3'.

Step 4: PRINT(z); becomes printf("int=%d,",z). Hence it prints 'int=4'.

Hence the output of the program is int=2, int=3, int=4.)

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5.     What will be the output of the program?

#include<stdio.h>
#define FUN(i, j) i##j

int main()
{
    int va1=10;
    int va12=20;
    printf("%d\n", FUN(va1, 2));
    return 0;
}

A.     10 ,   B.     20,  C.     1020  ,  D.     12
(NB:The following program will make you understand about ## (macro concatenation) operator clearly.)

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6.What will be the output of the program?

#include<stdio.h>
#define MAX(a, b) (a > b ? a : b)

int main()
{
    int x;
    x = MAX(3+2, 2+7);
    printf("%d\n", x);
    return 0;
}

A.     8,     B.     9 ,      C. 6 ,    D.     5
(NB:The macro MAX(a, b) (a > b ? a : b) returns the biggest value of the given two numbers.

Step 1 : int x; The variable x is declared as an integer type.

Step 2 : x = MAX(3+2, 2+7); becomes,

=> x = (3+2 > 2+7 ? 3+2 : 2+7)

=> x = (5 > 9 ? 5 : 9)

=> x = 9

Step 3 : printf("%d\n", x); It prints the value of variable x.

Hence the output of the program is 9.)

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7.    What will be the output of the program?

#include<stdio.h>
#define MIN(x, y) (x<y)? x : y;
int main()
{
    int x=3, y=4, z;
    z = MIN(x+y/2, y-1);
    if(z > 0)
        printf("%d\n", z);
    return 0;
}

A.     3    B.     4    C.     0    D.     No output

(NB:The macro MIN(x, y) (x<y)? x : y; returns the smallest value from the given two numbers.

Step 1: int x=3, y=4, z; The variable x, y, z are declared as an integer type and the variable x, y are initialized to value 3, 4 respectively.

Step 2: z = MIN(x+y/2, y-1); becomes,

=> z = (x+y/2 < y-1)? x+y/2 : y - 1;

=> z = (3+4/2 < 4-1)? 3+4/2 : 4 - 1;

=> z = (3+2 < 4-1)? 3+2 : 4 - 1;

=> z = (5 < 3)? 5 : 3;

The macro return the number 3 and it is stored in the variable z.

Step 3: if(z > 0) becomes if(3 > 0) here the if condition is satisfied. It executes the if block statements.

Step 4: printf("%d\n", z);. It prints the value of variable z. Hence the output of the program is 3)

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8.What will be the output of the program?

#include<stdio.h>
#define MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c)

int main()
{
    int x;
    x = MAX(3+2, 2+7, 3+7);
    printf("%d\n", x);
    return 0;
}

A.     5   , B.     9, C.     10    ,D.     3+7

(NB:The macro MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c) returns the biggest of given three numbers.

Step 1: int x; The variable x is declared as an integer type.

Step 2: x = MAX(3+2, 2+7, 3+7); becomes,

=> x = (3+2 >2+7 ? 3+2 > 3+7 ? 3+2 : 3+7: 2+7 > 3+7 ? 2+7 : 3+7)

=> x = (5 >9 ? (5 > 10 ? 5 : 10): (9 > 10 ? 9 : 10) )

=> x = (5 >9 ? (10): (10) )

=> x = 10

Step 3: printf("%d\n", x); It prints the value of 'x'.

Hence the output of the program is "10".)