Q: Which of the following is the correct usage of conditional operators used in C?

Option A: assignment statements are always return in paranthesis in the case of conditional operator. It should be a>b? (c=30):(c=40); Option B: it is syntatically wrong. Option D: syntatically wrong, it should be return(a>b ? a:b); Option C: it uses nested conditional operator, this is logic for finding greatest number out of three numbers.

Question 1

Which of the following is the correct order if calling functions in the below code? a = f1(23, 14) * f2(12/4) + f3();

Accepted Answer

Here, Multiplication will happen before the addition, but in which order the functions would be called is undefined. In an arithmetic expression the parenthesis tell the compiler which operands go with which operators but do not force the compiler to evaluate everything within the parenthesis first.

Question 2

Which of the following are unary operators in C? 1. ! 2. sizeof 3. ~ 4. &&

Accepted Answer

An operation with only one operand is called unary operation. Unary operators: ! Logical NOT operator. ~ bitwise NOT operator. sizeof Size-of operator. && Logical AND is a logical operator. Therefore, 1, 2, 3 are unary operators.

Question 3

Which of the following is the correct order of evaluation for the below expression? z = x + y * z / 4 % 2 - 1

Accepted Answer

C uses left associativity for evaluating expressions to break a tie between two operators having same precedence.

Question 4

Which of the following correctly shows the hierarchy of arithmetic operations in C?

Accepted Answer

Simply called as BODMAS (Bracket of Division, Multiplication, Addition and Subtraction). How Do I Remember ? BODMAS ! B - Brackets first O - Orders (ie Powers and Square Roots, etc.) DM - Division and Multiplication (left-to-right) AS - Addition and Subtraction (left-to-right)

Question 5

In which order do the following gets evaluated 1. Relational 2. Arithmetic 3. Logical 4. Assignment

Accepted Answer

2. Arithmetic operators: *, /, %, +, - 1. Relational operators: >, <, >=, <=, ==, != 3. Logical operators : !, &&, || 4. Assignment operators: =

Question 6

Which of the following is the correct usage of conditional operators used in C?

Accepted Answer

Option A: assignment statements are always return in paranthesis in the case of conditional operator. It should be a>b? (c=30):(c=40); Option B: it is syntatically wrong. Option D: syntatically wrong, it should be return(a>b ? a:b); Option C: it uses nested conditional operator, this is logic for finding greatest number out of three numbers.

Question 7

What will be the output of the program? #include<stdio.h>
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
", w, x, y, z);
    re…

Accepted Answer

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
", w, x, y, z); Hence the output is "1, 0, 1, 1".

Question 8

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 ", ch >= 'A'…

Accepted Answer

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 ", 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 9

What will be the output of the program? #include<stdio.h>
int main()
{
    int x=12, y=7, z;
    z = x!=4 || y == 2;
    printf("z=%d
", z);
    return 0;
}

Accepted Answer

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
", z); Hence the output of the program is "z=1".

Question 10

What will be the output of the program? #include<stdio.h>
int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i && ++j || ++k;
    printf("%d, %d, %d, %d
", i, j, k, m);
    return 0;
}

Accepted Answer

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
", 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 11

What will be the output of the program? #include<stdio.h>
int main()
{
    int i=2;
    int j = i + (1, 2, 3, 4, 5);
    printf("%d
", j);
    return 0;
}

Accepted Answer

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.

Question 12

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 ", num); return 0; }

Accepted Answer

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
", num); It prints the value of variable num '30' Step 3: Hence the output of the program is '30'

Question 13

What will be the output of the program? #include<stdio.h>
int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i || ++j && ++k;
    printf("%d, %d, %d, %d
", i, j, k, m);
    return 0;
}

Accepted Answer

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
", 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 14

What will be the output of the program? #include<stdio.h>
int main()
{
    static int a[20];
    int i = 0;
    a[i] = i  ;
    printf("%d, %d, %d
", a[0], a[1], i);
    return 0;
}

Accepted Answer

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 be 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
", a[0], a[1], i); Here a[0] = 0, a[1] = 0(because all staic variables are initialized to '0') and i = 0. Step 4: Hence the output is "0, 0, 0".

Question 15

Assuming, integer is 2 byte, What will be the output of the program? #include<stdio.h>

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

Accepted Answer

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 16

What will be the output of the program? #include<stdio.h>
int main()
{
    int x=4, y, z;
    y = --x;
    z = x--;
    printf("%d, %d, %d
", x, y, z);
    return 0;
}

Accepted Answer

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
", x, y, z); Hence it prints "2, 3, 3".

Question 17

What will be the output of the program? #include int main() { int x=55; printf("%d, %d, %d ", x<=55, x=40, x>=10); return 0; }

Accepted Answer

Step 1: int x=55; here variable x is declared as an integer type and initialized to '55'. Step 2: printf("%d, %d, %d ", 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 18

What will be the output of the program? #include<stdio.h>
int main()
{
    int a=100, b=200, c;
    c = (a == 100 || b > 200);
    printf("c=%d
", c);
    return 0;
}

Accepted Answer

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
", c); It prints the value of variable i=1 Hence the output of the program is '1'(one).

Question 19

What will be the output of the program? #include<stdio.h>
int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i && ++j && ++k;
    printf("%d, %d, %d, %d
", i, j, k, m);
    return 0;
}

Accepted Answer

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
", 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".

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