Java C Expression Quiz 1  Java C Expression Question and Answers
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Question 1 of 20
1. Question
Which of the following is the correct order of evaluation for the below expression?
z = x + y * z / 4 % 2 – 1Correct
C uses left associativity for evaluating expressions to break a tie between two operators having same precedence.
Incorrect
C uses left associativity for evaluating expressions to break a tie between two operators having same precedence.

Question 2 of 20
2. Question
In which order do the following gets evaluated
1. Relational
2. Arithmetic
3. Logical
4. AssignmentCorrect
2. Arithmetic operators: *, /, %, +, –
1. Relational operators: >, <, >=, <=, ==, !=
3. Logical operators : !, &&, 
4. Assignment operators: =Incorrect
2. Arithmetic operators: *, /, %, +, –
1. Relational operators: >, <, >=, <=, ==, !=
3. Logical operators : !, &&, 
4. Assignment operators: = 
Question 3 of 20
3. Question
Associativity has no role to play unless the precedence of operator is same.
Correct
Incorrect

Question 4 of 20
4. Question
Which of the following correctly shows the hierarchy of arithmetic operations in C?
Correct
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 (lefttoright)
AS – Addition and Subtraction (lefttoright)Incorrect
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 (lefttoright)
AS – Addition and Subtraction (lefttoright) 
Question 5 of 20
5. Question
Which of the following are unary operators in C?
1. !
2. sizeof
3. ~
4. &&Correct
An operation with only one operand is called unary operation.
Unary operators:
! Logical NOT operator.
~ bitwise NOT operator.
sizeof Sizeof operator.
&& Logical AND is a logical operator.Therefore, 1, 2, 3 are unary operators.
Incorrect
An operation with only one operand is called unary operation.
Unary operators:
! Logical NOT operator.
~ bitwise NOT operator.
sizeof Sizeof operator.
&& Logical AND is a logical operator.Therefore, 1, 2, 3 are unary operators.

Question 6 of 20
6. Question
The expression of the right hand side of  operators does not get evaluated if the left hand side determines the outcome.
Correct
Because, if a is nonzero then b will not be evaluated in the expression (a  b)
Incorrect
Because, if a is nonzero then b will not be evaluated in the expression (a  b)

Question 7 of 20
7. Question
Associativity of an operator is either Left to Right or Right to Left.
Correct
Yes, the associativity of an operator is either Left to Right or Right to Left.
Incorrect
Yes, the associativity of an operator is either Left to Right or Right to Left.

Question 8 of 20
8. Question
Which of the following is the correct usage of conditional operators used in C?
Correct
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.Incorrect
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 9 of 20
9. Question
In the expression a=b=5 the order of Assignment is NOT decided by Associativity of operators
Correct
The equal to = operator has RighttoLeft Associativity. So it assigns b=5 then a=b.
Incorrect
The equal to = operator has RighttoLeft Associativity. So it assigns b=5 then a=b.

Question 10 of 20
10. Question
Which of the following is the correct order if calling functions in the below code?
a = f1(23, 14) * f2(12/4) + f3();Correct
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.
Incorrect
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 11 of 20
11. Question
Will the expression *p = p be disallowed by the compiler?
Correct
Because, here even though the value of p is accessed twice it is used to modify two different objects p and *p
Incorrect
Because, here even though the value of p is accessed twice it is used to modify two different objects p and *p

Question 12 of 20
12. Question
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;
}Correct
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”.
Incorrect
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”.

Question 13 of 20
13. Question
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;
}Correct
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 nonzero 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”.
Incorrect
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 nonzero 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”.

Question 14 of 20
14. Question
Two different operators would always have different Associativity.
Correct
No, Two different operators may have same associativity.
Example:
Arithmetic operators like ++, — having RighttoLeft associativity.
Relational operators like >, >= also have LefttoRight associativity.Incorrect
No, Two different operators may have same associativity.
Example:
Arithmetic operators like ++, — having RighttoLeft associativity.
Relational operators like >, >= also have LefttoRight associativity. 
Question 15 of 20
15. Question
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;
}Correct
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”.
Incorrect
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 16 of 20
16. Question
Are the following two statement same?
1. a <= 20 ? (b = 30): (c = 30);
2. (a <=20) ? b : (c = 30);Correct
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;
}Incorrect
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;
} 
Question 17 of 20
17. Question
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;
}Correct
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”.Incorrect
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”. 
Question 18 of 20
18. Question
Assunming, integer is 2 byte, What will be the output of the program?
#include<stdio.h>
int main()
{
printf(“%x “, 2<<2);
return 0;
}Correct
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.Incorrect
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 19 of 20
19. Question
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 “, c);
return 0;
}Correct
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).Incorrect
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 20 of 20
20. Question
Every operator has an Associativity
Correct
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 leftassociative, rightassociative or nonassociative.
Incorrect
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 leftassociative, rightassociative or nonassociative.