Find Output Of Program

Question 1 :

What will be the output of the program?

#include
int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i && ++j && ++k;
    printf("%d, %d, %d, %d\n", i, j, k, m);
    return 0;
}


A). -2, 3, 1, 1
B). 2, 3, 1, 2
C). 1, 2, 3, 1
D). 3, 3, 1, 2
Answer : Option A

Explanation :

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i && ++j && ++k;
becomes m = -2 && 3 && 1;
becomes m = TRUE && TRUE; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j,k are increemented by '1'(one).

Hence the output is "-2, 3, 1, 1".


Question 2 :

Assunming, integer is 2 byte, What will be the output of the program?

#include
int main()
{
    printf("%x\n", -2<<2);
    return 0;
}


A). ffff
B). 0
C). fff8
D). Error
Answer : Option C

Explanation :

The integer value 2 is represented as 00000000 00000010 in binary system.
Negative numbers are represented in 2's complement method.
1's complement of 00000000 00000010 is 11111111 11111101 (Change all 0s to 1 and 1s to 0).
2's complement of 00000000 00000010 is 11111111 11111110 (Add 1 to 1's complement to obtain the 2's complement value).
Therefore, in binary we represent -2 as: 11111111 11111110.
After left shifting it by 2 bits we obtain: 11111111 11111000, and it is equal to "fff8" in hexadecimal system.


Question 3 :

What will be the output of the program?

#include
int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i || ++j && ++k;
    printf("%d, %d, %d, %d\n", i, j, k, m);
    return 0;
}


A). 2, 2, 0, 1
B). 1, 2, 1, 0
C). -2, 2, 0, 0
D). -2, 2, 0, 1
Answer : Option D

Explanation :

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.
Step 2: m = ++i || ++j && ++k; here (++j && ++k;) this code will not get executed because ++i has non-zero value.
becomes m = -2 || ++j && ++k;
becomes m = TRUE || ++j && ++k; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.
Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of variable 'i' only increemented by '1'(one). The variable j,k are not increemented.
Hence the output is "-2, 2, 0, 1".


Question 4 :

What will be the output of the program?

#include
int main()
{
    int x=12, y=7, z;
    z = x!=4 || y == 2;
    printf("z=%d\n", z);
    return 0;
}


A). z=0
B). z=1
C). z=4
D). z=2
Answer : Option B

Explanation :

Step 1: int x=12, y=7, z; here variable x, y and z are declared as an integer and variable x and y are initialized to 12, 7 respectively.
Step 2: z = x!=4 || y == 2;
becomes z = 12!=4 || 7 == 2;
then z = (condition true) || (condition false); Hence it returns 1. So the value of z=1.
Step 3: printf("z=%d\n", z); Hence the output of the program is "z=1".


Question 5 :

What will be the output of the program?

#include
int main()
{
    static int a[20];
    int i = 0;
    a[i] = i  ;
    printf("%d, %d, %d\n", a[0], a[1], i);
    return 0;
}


A). 1, 0, 1
B). 1, 1, 1
C). 0, 0, 0
D). 0, 1, 0
Answer : Option C

Explanation :

Step 1: static int a[20]; here variable a is declared as an integer type and static. If a variable is declared as static and it will ne automatically initialized to value '0'(zero).
Step 2: int i = 0; here vaiable i is declared as an integer type and initialized to '0'(zero).
Step 3: a[i] = i ; becomes a[0] = 0;
Step 4: printf("%d, %d, %d\n", a[0], a[1], i);
Here a[0] = 0, a[1] = 0(because all staic variables are initialized to '0') and i = 0.
Hence the output is "0, 0, 0".


Question 6 :

What will be the output of the program?

#include
int main()
{
    int i=4, j=-1, k=0, w, x, y, z;
    w = i || j || k;
    x = i && j && k;
    y = i || j &&k;
    z = i && j || k;
    printf("%d, %d, %d, %d\n", w, x, y, z);
    return 0;
}


A). 1, 1, 1, 1
B). 1, 1, 0, 1
C). 1, 0, 0, 1
D). 1, 0, 1, 1
Answer : Option D

Explanation :

Step 1: int i=4, j=-1, k=0, w, x, y, z; here variable i, j, k, w, x, y, z are declared as an integer type and the variable i, j, k are initialized to 4, -1, 0 respectively.
Step 2: w = i || j || k; becomes w = 4 || -1 || 0;. Hence it returns TRUE. So, w=1
Step 3: x = i && j && k; becomes x = 4 && -1 && 0; Hence it returns FALSE. So, x=0
Step 4: y = i || j &&k; becomes y = 4 || -1 && 0; Hence it returns TRUE. So, y=1
Step 5: z = i && j || k; becomes z = 4 && -1 || 0; Hence it returns TRUE. So, z=1.
Step 6: printf("%d, %d, %d, %d\n", w, x, y, z); Hence the output is "1, 0, 1, 1".


Question 7 :

What will be the output of the program?

#include
int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i && ++j || ++k;
    printf("%d, %d, %d, %d\n", i, j, k, m);
    return 0;
}


A). 1, 2, 0, 1
B). -3, 2, 0, 1
C). -2, 3, 0, 1
D). 2, 3, 1, 1
Answer : Option C

Explanation :

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i && ++j || ++k;
becomes m = (-2 && 3) || ++k;
becomes m = TRUE || ++k;.
(++k) is not executed because (-2 && 3) alone return TRUE.
Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j are increemented by '1'(one).
Hence the output is "-2, 3, 0, 1".


Question 8 :

What will be the output of the program?

#include
int main()
{
    int x=4, y, z;
    y = --x;
    z = x--;
    printf("%d, %d, %d\n", x, y, z);
    return 0;
}


A). 4, 3, 3
B). 4, 3, 2
C). 3, 3, 2
D). 2, 3, 3
Answer : Option D

Explanation :

Step 1: int x=4, y, z; here variable x, y, z are declared as an integer type and variable x is initialized to 4.
Step 2: y = --x; becomes y = 3; because (--x) is pre-decrement operator.
Step 3: z = x--; becomes z = 3;. In the next step variable x becomes 2, because (x--) is post-decrement operator.
Step 4: printf("%d, %d, %d\n", x, y, z); Hence it prints "2, 3, 3".


Question 9 :

What will be the output of the program?

#include
int main()
{
    int i=3;
    i = i++;
    printf("%d\n", i);
    return 0;
}


A). 3
B). 4
C). 5
D). 6
Answer : Option B

Question 10 :

What will be the output of the program?

#include
int main()
{
    int a=100, b=200, c;
    c = (a == 100 || b > 200);
    printf("c=%d\n", c);
    return 0;
}


A). c=100
B). c=200
C). c=1
D). c=300
Answer : Option C

Explanation :

Step 1: int a=100, b=200, c;
Step 2: c = (a == 100 || b > 200);
becomes c = (100 == 100 || 200 > 200);
becomes c = (TRUE || FALSE);
becomes c = (TRUE);(ie. c = 1)
Step 3: printf("c=%d\n", c); It prints the value of variable i=1
Hence the output of the program is '1'(one).


Question 11 :

What will be the output of the program?

#include
int main()
{
    int x=55;
    printf("%d, %d, %d\n", x<=55, x=40, x>=10);
    return 0;
}


A). 1, 40, 1
B). 1, 55, 1
C). 1, 55, 0
D). 1, 1, 1
Answer : Option A

Explanation :

Step 1: int x=55; here variable x is declared as an integer type and initialized to '55'.
Step 2: printf("%d, %d, %d\n", x<=55, x=40, x>=10);
In printf the execution of expressions is from Right to Left.
here x>=10 returns TRUE hence it prints '1'.
x=40 here x is assigned to 40 Hence it prints '40'.
x<=55 returns TRUE. hence it prints '1'.
Step 3: Hence the output is "1, 40, 1".


Question 12 :

What will be the output of the program?

#include
int main()
{
    int i=2;
    printf("%d, %d\n", ++i, ++i);
    return 0;
}


A). 3, 4
B). 4, 3
C). 4, 4
D). Output may vary from compiler to compiler
Answer : Option D

Explanation :

The order of evaluation of arguments passed to a function call is unspecified.
Anyhow, we consider ++i, ++i are Right-to-Left associativity. The output of the program is 4, 3.
In TurboC, the output will be 4, 3.
In GCC, the output will be 4, 4.


Question 13 :

What will be the output of the program?

#include
int main()
{
    int k, num=30;
    k = (num>5 ? (num <=10 ? 100 : 200): 500);
    printf("%d\n", num);
    return 0;
}


A). 200
B). 30
C). 100
D). 500
Answer : Option B

Explanation :

Step 1: int k, num=30; here variable k and num are declared as an integer type and variable num is initialized to '30'.
Step 2: k = (num>5 ? (num <=10 ? 100 : 200): 500); This statement does not affect the output of the program. Because we are going to print the variable num in the next statement. So, we skip this statement.
Step 3: printf("%d\n", num); It prints the value of variable num '30'
Step 3: Hence the output of the program is '30'


Question 14 :

What will be the output of the program?

#include
int main()
{
    char ch;
    ch = 'A';
    printf("The letter is");
    printf("%c", ch >= 'A' && ch <= 'Z' ? ch + 'a' - 'A':ch);
    printf("Now the letter is");
    printf("%c\n", ch >= 'A' && ch <= 'Z' ? ch : ch + 'a' - 'A');
    return 0;
}


A). The letter is a
Now the letter is A
B). The letter is A
Now the letter is a
C). Error
D). None of above
Answer : Option A

Explanation :

Step 1: char ch; ch = 'A'; here variable ch is declared as an character type an initialized to 'A'.

Step 2: printf("The letter is"); It prints "The letter is".

Step 3: printf("%c", ch >= 'A' && ch <= 'Z' ? ch + 'a' - 'A':ch);

The ASCII value of 'A' is 65 and 'a' is 97.

Here
=> ('A' >= 'A' && 'A' <= 'Z') ? (A + 'a' - 'A'):('A')
=> (TRUE && TRUE) ? (65 + 97 - 65) : ('A')
=> (TRUE) ? (97): ('A')
In printf the format specifier is '%c'. Hence prints 97 as 'a'.
Step 4: printf("Now the letter is"); It prints "Now the letter is".
Step 5: printf("%c\n", ch >= 'A' && ch <= 'Z' ? ch : ch + 'a' - 'A');
Here => ('A' >= 'A' && 'A' <= 'Z') ? ('A') : (A + 'a' - 'A')
=> (TRUE && TRUE) ? ('A') :(65 + 97 - 65)
=> (TRUE) ? ('A') : (97)
It prints 'A'

Hence the output is

The letter is a
Now the letter is A


Question 15 :

What will be the output of the program?

#include
int main()
{
    int i=2;
    int j = i + (1, 2, 3, 4, 5);
    printf("%d\n", j);
    return 0;
}


A). 4
B). 7
C). 6
D). 5
Answer : Option B

Explanation :

Because, comma operator used in the expression i (1, 2, 3, 4, 5). The comma operator has left-right associativity. The left operand is always evaluated first, and the result of evaluation is discarded before the right operand is evaluated. In this expression 5 is the right most operand, hence after evaluating expression (1, 2, 3, 4, 5) the result is 5, which on adding to i results into 7.