Question 1

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

Answer

Answer: B

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

Question 2

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

Answer

Answer: A

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

Question 3

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

Answer

Answer: A

Associativity 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 4

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;
}

Answer

Answer: B

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 5

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;
}

Answer

Answer: A

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

Questions for: Expressions