Java C Expression Quiz 1 | Java C Expression Question and Answers |

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

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

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)

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.

Because, if a is non-zero then b will not be evaluated in the expression (a || b)

Yes, the associativity of an operator is either Left to Right or Right to Left.

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.

The equal to = operator has Right-to-Left Associativity. So it assigns b=5 then a=b.

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.

Because, here even though the value of p is accessed twice it is used to modify two different objects p and *p

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 incremented by 1(one).

Hence the output is “-2, 3, 1, 1”.

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 (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 incremented by 1(one). The variable j,k are not increemented.

Hence the output is “-2, 2, 0, 1”.

No, Two different operators may have same associativity.

Example:
Arithmetic operators like ++, — having Right-to-Left associativity.
Relational operators like >, >= also have Left-to-Right associativity.

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”.

No, the expressions 1 and 2 are not same.

1. a <= 20 ? (b = 30) : (c = 30); This statement can be rewritten as,

if(a <= 20)
{
b = 30;
}
else
{
c = 30;
}
2. (a <=20) ? b : (c = 30); This statement can be rewritten as,

if(a <= 20)
{
//Nothing here
}
else
{
c = 30;
}

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 “, 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”.

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.

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).

Yes, Each and every operator has an associativity.

The associativity (or fixity) of an operator is a property that determines how operators of the same precedence are grouped in the absence of parentheses. Operators may be left-associative, right-associative or non-associative.