# C Programming Expressions Online Test, C Programming Test 2019

## C Programming Expressions Online Test

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C Programming Expressions Online Test. C Programming Question and Answers in English. **C Programming Expressions** Online mock test paper is free for all students and Very Helpful for Exam Preparation. C Programming Expressions Online Quiz. C Programming Online Mock test for Expressions Topic. Here we are providing** C Programming Expressions Online Test Series in English. Check C Programming **Mock Test Series 2019-2019.

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- Question 1 of 29
##### 1. Question

Which of the following is the correct order of evaluation for the below expression?

*z = x + y * z / 4 % 2 – 1*CorrectC uses left associativity for evaluating expressions to break a tie between two operators having same precedence.

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

- Question 2 of 29
##### 2. Question

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

CorrectSimply 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)

IncorrectSimply 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 3 of 29
##### 3. Question

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

CorrectOption 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.

IncorrectOption 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 4 of 29
##### 4. Question

Which of the following is the correct order if calling functions in the below code?

*a = f1(23, 14) * f2(12/4) + f3();*CorrectHere, 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.

IncorrectHere, 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 5 of 29
##### 5. Question

Which of the following are unary operators in C?

1. !

2. sizeof

3. ~

4. &&CorrectAn 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.

IncorrectAn 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 6 of 29
##### 6. Question

In which order do the following gets evaluated

1. Relational

2.Arithmetic

3.Logical

4.AssignmentCorrect2. Arithmetic operators:

**, /, %, +, –*

1. Relational operators:*>, <, >=, <=, ==, !=*

3. Logical operators :*!, &&, ||*

4. Assignment operators:*=*Incorrect2. Arithmetic operators:

**, /, %, +, –*

1. Relational operators:*>, <, >=, <=, ==, !=*

3. Logical operators :*!, &&, ||*

4. Assignment operators:*=* - Question 7 of 29
##### 7. 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\n", 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\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”.

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\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 8 of 29
##### 8. Question

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

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

CorrectThe 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.

IncorrectThe 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 9 of 29
##### 9. 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\n", 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 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”.

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 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 10 of 29
##### 10. 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\n", 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\n”, 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\n”, z);*Hence the output of the program is “z=1”. - Question 11 of 29
##### 11. 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\n", 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\n”, 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\n”, 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 12 of 29
##### 12. Question

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\n", w, x, y, z); return 0; }`

Correct**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”.Incorrect**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 13 of 29
##### 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\n", 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) || ++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”.

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) || ++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 14 of 29
##### 14. Question

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\n", x, y, z); return 0; }`

Correct**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”.Incorrect**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 15 of 29
##### 15. Question

What will be the output of the program?

`#include<stdio.h> int main() { int i=3; i = i++; printf("%d\n", i); return 0; }`

CorrectIncorrect - Question 16 of 29
##### 16. Question

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\n", 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\n”, 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\n”, c);*It prints the value of variable*i=1*

Hence the output of the program is ‘1’(one). - Question 17 of 29
##### 17. Question

What will be the output of the program?

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

Correct**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”.Incorrect**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 18 of 29
##### 18. Question

What will be the output of the program?

`#include<stdio.h> int main() { int i=2; printf("%d, %d\n", ++i, ++i); return 0; }`

CorrectThe 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.

IncorrectThe 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 19 of 29
##### 19. Question

What will be the output of the program?

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

Correct**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’Incorrect**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 20 of 29
##### 20. Question

What will be the output of the program?

`#include<stdio.h> 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;`

Correct**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 AIncorrect**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 21 of 29
##### 21. Question

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\n", j); return 0; }`

CorrectBecause, 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.IncorrectBecause, 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 22 of 29
##### 22. Question

Associativity has no role to play unless the precedence of operator is same.

CorrectAssociativity is only needed when the operators in an expression have the same precedence. Usually

*+*and*–*have the same precedence.Consider the expression

*7 – 4 + 2*. The result could be either*(7 – 4) + 2 = 5*or*7 – (4 + 2) = 1*. The former result corresponds to the case when*+*and*–*are left-associative, the latter to when*+*and*–*are right-associative.Usually the addition, subtraction, multiplication, and division operators are left-associative, while the exponentiation, assignment and conditional operators are right-associative. To prevent cases where operands would be associated with two operators, or no operator at all, operators with the same precedence must have the same associativity.

IncorrectAssociativity is only needed when the operators in an expression have the same precedence. Usually

*+*and*–*have the same precedence.Consider the expression

*7 – 4 + 2*. The result could be either*(7 – 4) + 2 = 5*or*7 – (4 + 2) = 1*. The former result corresponds to the case when*+*and*–*are left-associative, the latter to when*+*and*–*are right-associative.Usually the addition, subtraction, multiplication, and division operators are left-associative, while the exponentiation, assignment and conditional operators are right-associative. To prevent cases where operands would be associated with two operators, or no operator at all, operators with the same precedence must have the same associativity.

- Question 23 of 29
##### 23. Question

The expression of the right hand side of || operators doesn’t get evaluated if the left hand side determines the outcome.

CorrectBecause, if a is non-zero then b will not be evaluated in the expression*(a || b)*IncorrectBecause, if a is non-zero then b will not be evaluated in the expression*(a || b)* - Question 24 of 29
##### 24. Question

In the expression a=b=5 the order of Assignment is NOT decided by Associativity of operators

CorrectThe equal to

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

*=*operator has Right-to-Left Associativity. So it assigns*b=5*then*a=b*. - Question 25 of 29
##### 25. Question

Associativity of an operator is either Left to Right or Right to Left.

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

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

- Question 26 of 29
##### 26. Question

Are the following two statement same?

1. a <= 20 ? (b = 30): (c = 30);

2. (a <=20) ? b : (c = 30);CorrectNo, 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; }`

IncorrectNo, 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 27 of 29
##### 27. Question

Two different operators would always have different Associativity.

CorrectNo, 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.IncorrectNo, 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. - Question 28 of 29
##### 28. Question

Will the expression *p = p be disallowed by the compiler?

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

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

- Question 29 of 29
##### 29. Question

Every operator has an Associativity

CorrectYes, 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.

IncorrectYes, 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.