What is the output of the code?
❌ Nothing is printed
❌ i from A is 0
❌ i from A is 60
✅ i from A is 90
❌ Error
❌ None of the above
Explanation:
Class B is the subclass of class A. During creation of object of class the init function is called. Inside super().init() is used and it invokes the init function of super class A.
Then self.calcI(30) in turn performs the call of class method for which object is created, so calcI from class B is used. It means that 30 is multiplied by 3 and saved to "i". After all, the "i from A is 90" value is printed.
❌ Nothing is printed
❌ i from A is 0
❌ i from A is 60
✅ i from A is 90
❌ Error
❌ None of the above
Explanation:
Class B is the subclass of class A. During creation of object of class the init function is called. Inside super().init() is used and it invokes the init function of super class A.
Then self.calcI(30) in turn performs the call of class method for which object is created, so calcI from class B is used. It means that 30 is multiplied by 3 and saved to "i". After all, the "i from A is 90" value is printed.
What is the output of the code?
❌ Nothing
❌ Inbox Inbox
✅ Inbox Spam
❌ Spam Inbox
❌ Spam Spam
❌ Error
❌ None of the above
Explanation:
The output in this case is again 'Inbox' on one line and 'Spam' on another, because the print statement in the nested function finds the name in the enclosing function’s local scope, and the print at the end finds the variable in the global scope.
❌ Nothing
❌ Inbox Inbox
✅ Inbox Spam
❌ Spam Inbox
❌ Spam Spam
❌ Error
❌ None of the above
Explanation:
The output in this case is again 'Inbox' on one line and 'Spam' on another, because the print statement in the nested function finds the name in the enclosing function’s local scope, and the print at the end finds the variable in the global scope.
What will be the output of the code?
❌ The program has an error because there isn’t any function to return self.a
❌ The program has an error because b is private and display(self) is returning a private member
✅ The program has an error because b is private and hence can’t be printed
❌ The program runs fine and 1 is printed
Explanation:
Variables beginning with two underscores are said to be private members of the class and they can’t be accessed directly.
❌ The program has an error because there isn’t any function to return self.a
❌ The program has an error because b is private and display(self) is returning a private member
✅ The program has an error because b is private and hence can’t be printed
❌ The program runs fine and 1 is printed
Explanation:
Variables beginning with two underscores are said to be private members of the class and they can’t be accessed directly.
What is the output of the program?
❌ ['a', 'b', 'c']
❌ ['c', 'b', 'a']
❌ ['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b', 'c']
❌ ['c', 'b', 'a', 'c', 'b', 'a', 'c', 'b', 'a']
✅ []
❌ Error
❌ None of the mentioned
Explanation:
A expression list[listelements]*N where N is a integer appends N copies of list elements in the original list. If N is a negative integer or 0 output will be a empty list else if N is positive list elements will be added N times to the original list.
❌ ['a', 'b', 'c']
❌ ['c', 'b', 'a']
❌ ['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b', 'c']
❌ ['c', 'b', 'a', 'c', 'b', 'a', 'c', 'b', 'a']
✅ []
❌ Error
❌ None of the mentioned
Explanation:
A expression list[listelements]*N where N is a integer appends N copies of list elements in the original list. If N is a negative integer or 0 output will be a empty list else if N is positive list elements will be added N times to the original list.
What is the true statement about the code snippet?
❌ Class B inherits all the data fields of class A
❌ Class B needs an argument
❌ The data field ‘j’ cannot be accessed by object b
✅ Class B is inheriting class A but the data field ‘i’ in A cannot be inherited
Explanation:
Program ends with error 'B' object has no attribute 'i'.
Reason being that i is initiated with self thus making it a instantiate variable of that class which cannot be inherited by the above way.
❌ Class B inherits all the data fields of class A
❌ Class B needs an argument
❌ The data field ‘j’ cannot be accessed by object b
✅ Class B is inheriting class A but the data field ‘i’ in A cannot be inherited
Explanation:
Program ends with error 'B' object has no attribute 'i'.
Reason being that i is initiated with self thus making it a instantiate variable of that class which cannot be inherited by the above way.
What is the output of the code?
❌ {0: 1, 7: 0, 1: 1, 8: 0}
❌ {1: 1, 7: 2, 0: 1, 8: 1}
✅ {0: 0, 7: 0, 1: 1, 8: 1}
❌ KeyError
Explanation:
enumerate() will return a tuple, the loop will have x = (0, 0), (1, 1). Thus D[0] = 0, D[1] = 1, D[0 + 7] = D[7] = 0 and D[1 + 7] = D[8] = 1.
Note:
Dictionary is unordered, so the sequence of the key-value pair may differ in each output.
❌ {0: 1, 7: 0, 1: 1, 8: 0}
❌ {1: 1, 7: 2, 0: 1, 8: 1}
✅ {0: 0, 7: 0, 1: 1, 8: 1}
❌ KeyError
Explanation:
enumerate() will return a tuple, the loop will have x = (0, 0), (1, 1). Thus D[0] = 0, D[1] = 1, D[0 + 7] = D[7] = 0 and D[1 + 7] = D[8] = 1.
Note:
Dictionary is unordered, so the sequence of the key-value pair may differ in each output.
What is the output of the program?
❌ 0-0-5
❌ 1-2-10
❌ 2-1-5
✅ 2-0-5
❌ 3-1-10
❌ 3-2-6
❌ Error
Explanation:
First off, the * and % have the same operator precedence, so the expression is evaluated from left to right unless parentheses are present. The first expression evaluates to 8 % 3, which leaves a remainder of 2. The second expression is just evaluated left to right since * and % have the same operator precedence, and it reduces to 6 % 3, which is 0. The last expression reduces to 5 * 1, which is 5. Therefore, the output on line 6 is 2-0-5.
❌ 0-0-5
❌ 1-2-10
❌ 2-1-5
✅ 2-0-5
❌ 3-1-10
❌ 3-2-6
❌ Error
Explanation:
First off, the * and % have the same operator precedence, so the expression is evaluated from left to right unless parentheses are present. The first expression evaluates to 8 % 3, which leaves a remainder of 2. The second expression is just evaluated left to right since * and % have the same operator precedence, and it reduces to 6 % 3, which is 0. The last expression reduces to 5 * 1, which is 5. Therefore, the output on line 6 is 2-0-5.