## Programming-for-Output-Problems

Question 1 |

Consider the following C program:

#include <stdio.h> int main () { int arr [] = {1,2,3,4,5,6,7,8,9,0,1,2,5}, *ip = arr+4; printf ("%d\n", ip[1]); return 0; }

The number that will be displayed on execution of the program is _____.

5 | |

6 | |

7 | |

8 |

We know that arr is a pointer to arr[ ] & hence arr+4 is pointer to 4

^{th}index of array (starting from 0 to 4).

Now *ip is a pointer of int type pointing to memory location 108, which is part of arr.

Hence, when we will print ip[1] it will be equivalent to *(ip+1).

Address of ip will be incremented by 1 & value inside 110 will be printed.

Question 2 |

Consider the following C program:

#include <stdio.h> int jumble (int x, int y) { x = 2 * x + y ; return x ; } int main ( ) { int x=2, y=5 ; y = jumble (y, x) ; x = jumble (y, x) ; printf ("%d \n", x) ; return 0 ; }

The value printed by the program is ______.

26 | |

67 | |

25 | |

13 |

#include

int jumble(int x, int y)

{

printf("Inside jumble : 2*%d + %d\n", x,y);

x = 2*x +y;

return x;

}

int main()

{

int x=2, y=5;

printf("Initial x=%d, y=%d\n",x,y);

printf("1st jumble call : jumble(%d,%d)\n",y,x);

y = jumble(y,x);

printf("Value of y after 1st jumble = %d\n", y);

printf("2

^{nd}jumble call: jumble(%d,%d)\n", y,x);

x = jumble(y,x);

printf("Value of x after 2nd jumble = %d\n", x);

printf("Final : %d\n", x);

return 0;

}

////////////////////////////////////OUTPUT

Initial x=2, y=5

1

^{st}jumble call: jumble(5,2)

Inside jumble : 2*5 + 2

Value of y after 1

^{st}jumble = 12

2

^{nd}jumble call: jumble(12,2)

Inside jumble : 2*12 + 2

Value of x after 2nd jumble = 26

Final : 26

Question 3 |

Consider the following C program:

#include <stdio.h> int main() { int a[] = {2, 4, 6, 8, 10} ; int i, sum = 0, *b = a + 4 ; for (i = 0; i < 5; i++) sum = sum + (*b - i) - *(b - i) ; printf ("%d\n", sum) ; return 0 ; }

The output of the above C program is _____.

3 | |

7 | |

11 | |

10 |

#include

int main()

{

int a[] = {2,4,6,8,10};

int i, sum = 0, *b = a+4;

for(i=0; i<5; i++)

{ printf("*b, (*b-i): %d , %d\n",*b, (*b-i) );

printf("*(b-i): %d\n",*(b-i) );

printf("sum = %d + %d - %d\n",sum, (*b-i),*(b-i));

sum = sum + (*b-i) - *(b-i);

printf("sum = %d\n", sum);

}

printf("%d\n", sum);

return 0;

}

//////////////////////////////OUTPUT

*b, (*b-i): 10 , 10

*(b-i): 10

sum = 0 + 10 - 10

sum = 0

*b, (*b-i): 10 , 9

*(b-i): 8

sum = 0 + 9 - 8

sum = 1

*b, (*b-i): 10 , 8

*(b-i): 6

sum = 1 + 8 - 6

sum = 3

*b, (*b-i): 10 , 7

*(b-i): 4

sum = 3 + 7 - 4

sum = 6

*b, (*b-i): 10 , 6

*(b-i): 2

sum = 6 + 6 - 2

sum = 10

10

Question 4 |

Consider the following C program:

#include <stdio.h> int r() { static int num=7 ; return num-- ; } int main () { for (r(); r (); r()) printf ("%d", r()); return 0 ; }

Which one of the following values will be displayed on execution of the programs?

41 | |

63 | |

52 | |

630 |

#include

int r()

{

int x;

static int num=7;

x =num--;

printf("num--: %d\n",x);

return x;

}

int main()

{

for(r(); r(); r())

{

printf("%d\n", r());

}

return 0;

}

//////////////////////////////OUTPUT

num--: 7

num--: 6

num--: 5

5

num--: 4

num--: 3

num--: 2

2

num--: 1

num--: 0

Question 5 |

Consider the following C function.

void convert (int n) { if (n < 0) printf ("%d", n); else { convert (n/2); printf ("%d", n%2); } }

Which one of the following will happen when the function convert is called with any positive integer n as argument?

It will print the binary representation of n in the reverse order and terminate. | |

It will not print anything and will not terminate. | |

It will print the binary representation of n and terminate. | |

It will print the binary representation of n but will not terminate. |

Sequence of function calls

Convert(6)

Convert(3)

Convert(1)

Convert(0)

:

Convert(0)

:

:

It will not terminate and never produce any output.

Note:

There is no instruction which stops the loop.

Question 6 |

Consider the following C program:

#include <stdio.h> int main () { float sum = 0.0, j = 1.0, i = 2.0; while (i/j > 0.0625) { j = j + j; sum = sum + i/j; printf ("%f \n", sum); } return 0; }

The number of times the variable sum will be printed, when the above program is executed, is ______.

5 | |

2 | |

7 | |

10 |

#include

int main()

{

float sum= 0.0, j=1.0, i=2.0;

while(i/j > 0.0625)

{

j = j+j;

sum = sum+i/j;

printf("%f\n",sum);

}

return 0;

}

//////////////////////////////////OUTPUT

1.000000

1.500000

1.750000

1.875000

1.937500

Question 7 |

Computes the LCM of two numbers | |

Divides the larger number by the smaller number | |

Computes the GCD of two numbers | |

Finds the smaller of two numbers |

→ The procedure is to subtract smaller number from larger. So that we can reduce larger number then doesn’t change the value of GCD.

→ if we performing number of iterations based on condition the larger of two numbers will end up with GCD.

Question 8 |

z2 | |

z1z2 | |

Compilation error | |

None of these |

Question 9 |

0 | |

99999999 | |

1 | |

unpredictable |

Question 10 |

person.name +2 | |

kd → (name +2 ) | |

*((*kd).name + 2 ) | |

either (a) or (b), but not (c) |

We have to write *((*kd).name+2).

Note: They are specifically asked about a value stored at that location of the string by using "*" outside the brackets. Otherwise, option A and B would be correct.

Question 11 |

0-9 lines of output | |

10-19 lines of output | |

20-29 lines of output | |

More than 29 lines of output |

times the loop executes. Since i is constant, we need to see the growth of j only.

→ Let the initial value of j be denoted by J1 and the subsequent values by J

_{n}for n= 2, 3,... so that J denotes a progression of j. We see that J

_{n}= 2J

_{n-1};

→ J

_{1}= 2, which gives J

_{n}= 2

^{n-1};

n = 1, 2, ...

→ The loop will execute as long as i/j=2.0/2

^{n-1}>0.001 which gives n<3log

_{2}10+2 (or) n<11.97.

→ Thus the loop will execute 11 times which is equivalent to say there will be 11 lines of output.

Question 12 |

910 | |

920 | |

870 | |

900 |

Step-2: We pass address of number as parameters in the square function. As per the concept of function overloading in C++ the second square function will be executed.

Step-3: --(*y) will be executed first since decrement operator has the higher precedence than multiplication.

*x = (*x) * --(*y);

*x = 30 * 29;

*x = 870

Note: x is a pointer variable and it holds the address of the variable number.

Question 13 |

**Consider the following C function:**

int f(int n)

{

static int i = 1;

if(n >= 5) return n;

n = n+i;

i++;

return f(n);

}

The value returned by f(1) is

5
| |

6 | |

7 | |

8 |

first line of f() is executed only once.

Depending upon the value “n” and “i”, the function f() will until n>=5.

You can find the tracing of function f() from the below Execution of f(1)

i = 1

n = 2

i = 2

Call f(2)

i = 2

n = 4

i = 3

Call f(4)

i = 3

n = 7

i = 4

Call f(7)

since n >= 5 return n(7) with return value = 7

Question 14 |

**Consider the following code segment**

for (int k=0; k<20; k=k+2)

{

if (k % 3 == 1)

system.out.print(k+ " ")

}

What is printed as a result of executing the code segment?

4 16 | |

4 10 16 | |

0 6 12 18 | |

1 4 7 10 13 16 19 |

k = 0 % 3 = 0

k = 2 % 3 = 2

k = 4 % 3 = 1 // prints 4

k = 6 % 3 = 0

k = 8 % 3 = 2

k = 10 % 3 = 1 // prints 10

k = 12 % 3 = 0

k = 14 % 3 = 2

k = 16 % 3 = 1 // prints 16

k = 18 % 3 = 0

So, Output is 4 10 16

Question 15 |

5 5 5 | |

5 5 junk | |

5 junk junk | |

Compile time error |

The same rule applies in **m.

Output is 5 5 5

Question 16 |

1 | |

320 | |

64 | |

Compilation error |

Printf (“%d”, *ptr) ; /* Here *ptr is referring a single character */

Printf (“%d”, *ptr) ;

Output: 64 /* @ ASCII value is 64 */

Note: Assume short a is 2 byte integer.

Question 17 |

⌊log _{2}n⌋*n | |

n | |

⌊log _{2}n⌋ | |

⌊log _{2}n⌋+1 |

Let us consider n=6, then

1<=6 (correct)

2<=6 (correct)

4<=6 (correct)

8<=6 (False)

4 comparisons required

Option A:

⌊log n⌋+1

⌊log 6⌋+1

3+1=4 (correct)

Option B:

n=6 (False)

Option C:

⌊log n⌋

⌊log 6⌋=3 (False)

Option D:

⌊log

_{2}n⌋+1

⌊log

_{2}6⌋+1=2+1=3 (False)

Question 18 |

What is the output of the following C code?

#include

int main()

{

int index;

for(index=1; index<=5; index++)

{

printf("%d", index);

if (index==3)

continue;

}

}

1245 | |

12345 | |

12245 | |

12354 |

In the given code, there are no statements after continue statement , So it won’t effect on the output.

The loop will executes for five iterations,For each iteration it will print corresponding value i.e; 12345.

Question 19 |

#include

#include

void main()

{

double pi = 3.1415926535;

int a = 1;

int i;

for(i=0; i < 3; i++)

if(a = cos(pi * i/2) )

printf("%d ",1);

else printf("%d ", 0);

}

What would the program print?

000 | |

010 | |

101 | |

111 |

→For a given i = 0:

a = cos(pi * 0/2) [ PI*0 is 0]

a = cos(0) = 1, if condition true and it will print 1

→For a given i = 1

a = cos (pi/2) [pi*1 is pi]

a=cos(1.57075) whose value approximately equal to zero

a = 0, Here the condition is false then else part will execute and it will print 0

→For i = 2

a = cos(pi) [pi*2/2 is nothing but pi]

a = -1, ,Here also condition is true and it will execute if part and it will print “1”

→Finally it will print 101.

Question 20 |

#include <stdio.h>

int main (void)

{

int shifty;

shifty = 0570;

shifty = shifty >>4;

shifty = shifty <<6;

printf("the value of shifty is %o",shifty);

}

the value of shifty is 1500 | |

the value of shifty is 4300 | |

the value of shifty is 5700 | |

the value of shifty is 2700 |

(0570)8 = (000 101 111 000)

_{2}(Converting octal number into binary number)

shifty = shifty >>4 ( >> right shift operator where we need to shift the 4 bits towards right side means discard last four bits )

After right shifting 4 bits the shifty consists of the following digits

shifty = (000 000 010 111)

_{2}

shifty = shifty <<6( << left shift operator where we need to shift the 6 bits towards left side means add si bits to the end of the binary number. So the binary number becomes as follows shifty = (010 111 000 000)

_{2}

= (2700)

_{8}

Question 21 |

y=y+1;

z=x+y;

x=x+1

z = x + y + 2; | |

z = (x++) + (++y); | |

z = (x++) + (y++); | |

z = (x++) + (++y) + 1; |

y=y+1;

z=x+y;

x=x+1

First statement: “y” value is incremented by 1

Second statement : that incremented value is added to x and stored into “z”

Third statement : “y” value is incremented by 1

z = (x++) + (++y);

X++ post-increment , so it will perform action which is addition and later the value of “x” is incremented .

++y pre-increment , here first value “y” is incremented and updated value is added to value x.

Finally the result will store into z.

The sequence of operations you can find from the below statements

z=x+y; //z = x + (++y)

x=x+1 //z = (x++) + (++y)

Question 22 |

#include<stdio.h>

float i=2.0;

float j=1.0;

float sum = 0.0;

main()

{

while (i/j > 0.001)

{

j+=j;

sum=sum+(i/j);

printf("%f\n", sum);

}

}

8 | |

9 | |

10 | |

11 |

while (1.000000 > 0.001)

{

j=2.0

sum=0+1.000000;

printf("%f\n",sum); /* It will print 1.000000 */

}

Iteration-2: 1.500000

Iteration-3: 1.750000

Iteration-4: 1.875000

Iteration-5: 1.937500

Iteration-6: 1.968750

Iteration-7: 1.984375

Iteration-8: 1.992188

Iteration-9: 1.996094

Iteration-10: 1.998047

Iteration-11: 1.999023

The program will terminate after 11th iteration. So, it print 11 lines.

Question 23 |

int main()

{

static int x[ ]={1,2,3,4,5,6,7,8}

int i;

for(i=2;i<6;++i)

x[x[i]]=x[i];

for(i=0;i<8;++i)

printf(“%d”,x[i]);

Return 0;

}

12244668 | |

11335578 | |

12335578
| |

12343676 |

Question 24 |

void foo(int x, int y)

{

x+=y;

y+=x;

}

main()

{

int x=5.5;

foo(x,x);

}

What is the final value of x in both call by value and call by reference, respectively?

5 and 16 | |

5 and 12 | |

5 and 20 | |

12 and 20 |

__Call by Value:__

The call by value method of passing arguments to a function copies the actual value of an argument into the formal parameter of the function. In this case, changes made to the parameter inside the function have no effect on the argument.

foo(x,x) ---->foo(5,5)

Modification of x and y values won’t effect in the main program, So still x will hold value “5” because the data type is integer.

__Call by Reference:__

The call by reference method of passing arguments to a function copies the address of an argument into the formal parameter. Inside the function, the address is used to access the actual argument used in the call. It means the changes made to the parameter affect the passed argument.

Function call foo(x,x)---foo(5,5) calls the function

void foo(int x, int y)

{

x+=y;

y+=x;

}

Here x and y variables, both pointing to same variable “x” in the main function.

So, if any modification done in the function it automatically reflect in the main function.

x+=y means x=x+y=5+5=10 // The variable “x” of main function will hold value 10 which used by both local variables of x and y

y+=x means u=y+x= 10+10=20 // Y value is 10 because , it point to variable ”x” in the main function.

Question 25 |

struct

{

short s[5];

union

{

float y;

long z;

}u;

}t;

Assume that the objects of the type short, float and long occupy 2 bytes, 4 bytes and 8 bytes, respectively. The memory requirement for variable t, ignoring alignment consideration, is

22 bytes | |

18 bytes | |

14 bytes | |

10 bytes |

Question 26 |

**Consider the following program fragment**

if(a > b)

if(b > c)

s1;

else

s2;

s2 will be executed if

a <= b | |

b > c | |

b >= c and a <= b | |

a > b and b <= c |

We can write code segment is like this because to avoid “dangling if”

if(a>b)

{

if(b>c)

{ s1; }

else

{ s2; }

}

→ To execute s2, it (a>b) should be TRUE. And it (b>c) should be FALSE.

Hence, Option-D is correct

Question 27 |

assigns 3 to a[5] | |

assigns 4 to a[5] | |

assigns 4 to a[4] | |

what is assigned is compiler dependent |

Question 28 |

main()

{

inc();

inc();

inc();

}

inc()

{

static int x;

printf("%d", ++x);

}

prints 012 | |

prints 123 | |

prints 3 consecutive, but unpredictable numbers | |

prints 111 |

Static storage class by default value is 0.

Step-1: We are calling inc( ) in main function. So, it will enter into inc() function.

Here, we are using pre increment function, so it replaced 0 with 1.

Step-2: Second inc() in main function. So, it will enter into inc() function.

Here, we are using pre increment function, so it replaced 1 with 2.

Step-3: Third inc() in main function. So, it will enter into inc() function.

Here, we are using pre increment function, so it replaced 2 with 3.

Note: The static storage class instructs the compiler to keep a local variable in existence during the life-time of the program instead of creating and destroying it each time it comes into and goes out of scope.

Question 29 |

i=6720;

j=4;

while (i%j==0)

{

i=i/j;

j=j+1;

}

On termination j will have the value

4 | |

8 | |

9 | |

6720 |

__Iteration 1:__

i=6720

j=4

i%j==0 which is 6720%4 ==0 condition is true then

i=i/j=6720/4=1680

j=j+1=5

__Iteration-2:__

i=1630

j=5

i%j==0 which is 1680%5 ==0 condition is true then

i=i/j=1630/5=336

j=j+1=6

__Iteration-3:__i=336

j=6

i%j==0 which is 336%6 ==0 condition is true then

i=i/j=336/6=56

j=j+1=7

__Iteration-4:__

i=56

j=7

i%j==0 which is 56%7 ==0 condition is true then

i=i/j=56/7=8

j=j+1=8

__Iteration-5:__

i=8

j=8

i%j==0 which is 8%8 ==0 condition is true then

i=i/j=8/8=1

j=j+1=9

__Iteration-6:__

i=1

j=9

i%j==0 which is 1%9 ==0 condition is false.

At this iteration, the “j” value is 9.

Question 30 |

int a, *b=&a, **c=&b;

The following program fragment

a=4;

**c=5;

does not change the value of a | |

assigns address of c to a | |

assigns the value of b to a | |

assigns 5 to a |

Ordinary variable -a // value will be stored

Pointer variable -b // address of variable will be stored

Pointer to pointer variable(double pointer) -c // address of pointer variable will stored.

a=4 means storing/assigning value to “a”.

**c means value at(value at(c)) =value at(value at(&b)) (c holds address of pointer “b”)

=value at(&a) (b holds the address of “a”)

Memory location of variable “a”, value 5 is stored/assigned into the variable.

Question 31 |

main()

{

static int x[] = {1,2,3,4,5,6,7,8}

int i;

for (i=2; i<6; ++i)

x[x[i]]=x[i];

for (i=0; i<8; ++i)

printf("%d", x[i]);

}

1 2 3 3 5 5 7 8 | |

1 2 3 4 5 6 7 8 | |

8 7 6 5 4 3 2 1 | |

1 2 3 5 4 6 7 8 |

For loop will execute for the i value 2,3,4,5

For i = 2, x[x[2]] = x[2]

= x[3] = 3 // since x[2] = 3

The array elements representation are

Question 32 |

int n=17; | |

char c=99; | |

float f=(float)99.32; | |

#include |

Option B: General declaration of character variable.

Option C: It is type casting. But given cast also same type.

Option D: We have to include a file otherwise it treated as compiler error.

Question 33 |

for (i=0; i<10; ++i)

printf("%d", i&1);

prints:

0101010101 | |

0111111111 | |

0000000000 | |

1111111111 |

“&” is bitwise AND operator.

For a given i=0 , the value is 0&1 is 0000 & 0001 which is 0

i=1, 0001 & 0001 which is 1

i=2 , 0010 & 0001 which is 0

i=3 0011 & 0001 which is 1

i=4 , 0100 & 0001 which is 0

i=5 0101 & 0001 which is 1

i=6 , 1100 & 0001 which is 0

i=7 0111 & 0001 which is 1

i=8 , 1000 & 0001 which is 0

i=9 1001 & 0001 which is 1

So the output is 0101010101(from i=0 to 9)

Question 34 |

main()

{

static int a=3;

printf("%d",a--);

if(a)

main();

}

3 | |

3 2 1 | |

3 3 3 | |

Program will fall in continuous loop and print 3 |

Step-2: In printf( ) we are given post decrement. So, it will print 3 then decrements.

Step-3: if(2) then we are calling main() function, a=2 because of static otherwise it will be 3 only.

Step-4: It will print 2 and next iteration will print 1.