UGC NET CS 2016 July- paper-2
March 15, 2024UGC NET CS 2016 July- paper-2
March 15, 2024UGC NET CS 2016 July- paper-2
| Question 11 |
Given i = 0, j = 1, k = –1 x = 0.5, y = 0.0 What is the output of the following expression in C language ?
x * y < i + j || k
x * y < i + j || k
| -1 | |
| 0 | |
| 1 | |
| 2 |
Question 11 Explanation:
x * y < i + j || k
Step-1: Evaluate x * y because multiplication has more priority than remaining operators
x * y→ 0
Step-2: i + j is 1
Step-3: (x*y) < (i+j) is 1. Because relational operators only return 1(TRUE) or 0(FALSE).
Step-4: ((x*y) < (i+j)) || k is logical OR operator.
1 || -1 will returns 1.
Note: The precedence is ((x*y) < (i+j)) || k
Step-1: Evaluate x * y because multiplication has more priority than remaining operators
x * y→ 0
Step-2: i + j is 1
Step-3: (x*y) < (i+j) is 1. Because relational operators only return 1(TRUE) or 0(FALSE).
Step-4: ((x*y) < (i+j)) || k is logical OR operator.
1 || -1 will returns 1.
Note: The precedence is ((x*y) < (i+j)) || k
Correct Answer: C
Question 11 Explanation:
x * y < i + j || k
Step-1: Evaluate x * y because multiplication has more priority than remaining operators
x * y→ 0
Step-2: i + j is 1
Step-3: (x*y) < (i+j) is 1. Because relational operators only return 1(TRUE) or 0(FALSE).
Step-4: ((x*y) < (i+j)) || k is logical OR operator.
1 || -1 will returns 1.
Note: The precedence is ((x*y) < (i+j)) || k
Step-1: Evaluate x * y because multiplication has more priority than remaining operators
x * y→ 0
Step-2: i + j is 1
Step-3: (x*y) < (i+j) is 1. Because relational operators only return 1(TRUE) or 0(FALSE).
Step-4: ((x*y) < (i+j)) || k is logical OR operator.
1 || -1 will returns 1.
Note: The precedence is ((x*y) < (i+j)) || k
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