Jai solanki
| Description: Ssc | |
| Number of Questions: 37 | |
| Created by: Sunita singh | |
| Tags: Ch 1 |
Overriding and Overloading is for ?
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Methods and variables
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class
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variables
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Methods
To answer this question, the user needs to know the basic concepts of object-oriented programming (OOP) and the differences between method overriding and method overloading.
Option A: Methods and variables - This option is partially correct. Overriding and overloading both apply to methods in OOP, not to variables. Therefore, this option is only correct for the first part of the question.
Option B: Class - This option is not correct. Overriding and overloading are not for classes, but for methods within classes.
Option C: Variables - This option is not correct. Overriding and overloading are not for variables, but for methods within classes.
Option D: Methods - This option is correct. Overriding and overloading are techniques used for methods in OOP. Method overriding means providing a new implementation of a method in a subclass that is already defined in its superclass. Method overloading means defining multiple methods with the same name in the same class, but with different parameters.
Therefore, the correct answer is:
The answer is: D. Methods
Overriding and Overloading is for ?
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Methods and variables
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class
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variables
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Methods
To answer this question, the user needs to know the basic concepts of object-oriented programming (OOP) and the differences between method overriding and method overloading.
Option A: Methods and variables - This option is partially correct. Overriding and overloading both apply to methods in OOP, not to variables. Therefore, this option is only correct for the first part of the question.
Option B: Class - This option is not correct. Overriding and overloading are not for classes, but for methods within classes.
Option C: Variables - This option is not correct. Overriding and overloading are not for variables, but for methods within classes.
Option D: Methods - This option is correct. Overriding and overloading are techniques used for methods in OOP. Method overriding means providing a new implementation of a method in a subclass that is already defined in its superclass. Method overloading means defining multiple methods with the same name in the same class, but with different parameters.
Therefore, the correct answer is:
The answer is: D. Methods
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class, if, void, long, Int, continue
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goto, instanceof, native, finally, default, throws
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try, virtual, throw, final, volatile, transient
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strictfp, constant, super, implements, do
All the words in option B are among the 49 Java keywords. Although goto reserved as a keyword in Java, goto is not used and has no function.Option A is wrong because the keyword for the primitive int starts with a lowercase i.Option C is wrong because "vi
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class, if, void, long, Int, continue
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goto, instanceof, native, finally, default, throws
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try, virtual, throw, final, volatile, transient
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strictfp, constant, super, implements, do
All the words in option B are among the 49 Java keywords. Although goto reserved as a keyword in Java, goto is not used and has no function.Option A is wrong because the keyword for the primitive int starts with a lowercase i.Option C is wrong because "vi
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int [] myList = {"1", "2", "3"};
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int [] myList = (5, 8, 2);
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int myList [] [] = {4,9,7,0};
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int myList [] = {4, 3, 7};
To legally declare, construct, and initialize an array, the user needs to know the syntax of creating an array and how to initialize it with values.
Option A is incorrect because it uses double quotes for the array initialization, which is used for string values. For integer values, we don't need to use quotes.
Option B is incorrect because it uses parentheses instead of curly brackets for array initialization.
Option C is incorrect because it declares a two-dimensional array but only initializes with one-dimensional values.
Option D is correct because it declares, constructs, and initializes a one-dimensional integer array with the values 4, 3, and 7.
Therefore, the answer is:
The Answer is: D
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int [] myList = {"1", "2", "3"};
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int [] myList = (5, 8, 2);
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int myList [] [] = {4,9,7,0};
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int myList [] = {4, 3, 7};
To legally declare, construct, and initialize an array, the user needs to know the syntax of creating an array and how to initialize it with values.
Option A is incorrect because it uses double quotes for the array initialization, which is used for string values. For integer values, we don't need to use quotes.
Option B is incorrect because it uses parentheses instead of curly brackets for array initialization.
Option C is incorrect because it declares a two-dimensional array but only initializes with one-dimensional values.
Option D is correct because it declares, constructs, and initializes a one-dimensional integer array with the values 4, 3, and 7.
Therefore, the answer is:
The Answer is: D
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method
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native
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subclasses
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reference
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method
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native
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subclasses
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reference
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interface
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string
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Float
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unsigned
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interface
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string
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Float
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unsigned
Which one of the following will declare an array and initialize it with five numbers?
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Array a = new Array(5);
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int [] a = {23,22,21,20,19};
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int a [] = new int[5];
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int [5] array;
To declare an array and initialize it with five numbers, the user needs to know the syntax for declaring and initializing an array in Java.
Option A is incorrect because the Array class does not have a constructor that takes an integer as an argument, and thus this statement will not create an array of five elements.
Option B is correct because it initializes an integer array a with 5 elements and assigns the values 23, 22, 21, 20, and 19 to the array.
Option C is also correct because it declares an integer array a with 5 elements, but it does not initialize the array with any values.
Option D is incorrect because the syntax int [5] array is not valid in Java. The correct syntax should be int[] array = new int[5] which creates an integer array array with 5 elements and initializes all elements to 0.
Therefore, the correct answer is:
The Answer is: B. int [] a = {23,22,21,20,19};
Which one of the following will declare an array and initialize it with five numbers?
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Array a = new Array(5);
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int [] a = {23,22,21,20,19};
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int a [] = new int[5];
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int [5] array;
To declare an array and initialize it with five numbers, the user needs to know the syntax for declaring and initializing an array in Java.
Option A is incorrect because the Array class does not have a constructor that takes an integer as an argument, and thus this statement will not create an array of five elements.
Option B is correct because it initializes an integer array a with 5 elements and assigns the values 23, 22, 21, 20, and 19 to the array.
Option C is also correct because it declares an integer array a with 5 elements, but it does not initialize the array with any values.
Option D is incorrect because the syntax int [5] array is not valid in Java. The correct syntax should be int[] array = new int[5] which creates an integer array array with 5 elements and initializes all elements to 0.
Therefore, the correct answer is:
The Answer is: B. int [] a = {23,22,21,20,19};
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public double methoda();
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public final double methoda();
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static void methoda(double d1);
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protected void methoda(double d1);
To solve this question, the user needs to understand the syntax and rules for declaring methods within an interface in Java.
In Java, an interface is a collection of abstract methods, which means the methods declared within an interface do not have a body. The purpose of an interface is to define a contract that classes can implement, specifying the methods they must provide.
Now, let's go through each option and explain why it is right or wrong:
A. public double methoda(); This option is a valid declaration within an interface definition. It declares a public method named "methoda" that returns a double value. Since interfaces only contain abstract methods, there is no need to provide a method body.
B. public final double methoda(); This option is not a valid declaration within an interface definition. The "final" keyword cannot be used to modify a method declaration within an interface. The "final" keyword is used to indicate that a method cannot be overridden by a subclass.
C. static void methoda(double d1); This option is not a valid declaration within an interface definition. The "static" keyword cannot be used to modify a method declaration within an interface. Static methods belong to the class itself, not an instance of the class.
D. protected void methoda(double d1); This option is not a valid declaration within an interface definition. The "protected" keyword cannot be used to modify a method declaration within an interface. Protected methods are accessible within the same package and by subclasses.
Therefore, the valid declaration within an interface definition is:
The Answer is: A. public double methoda();
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public double methoda();
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public final double methoda();
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static void methoda(double d1);
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protected void methoda(double d1);
To solve this question, the user needs to understand the syntax and rules for declaring methods within an interface in Java.
In Java, an interface is a collection of abstract methods, which means the methods declared within an interface do not have a body. The purpose of an interface is to define a contract that classes can implement, specifying the methods they must provide.
Now, let's go through each option and explain why it is right or wrong:
A. public double methoda(); This option is a valid declaration within an interface definition. It declares a public method named "methoda" that returns a double value. Since interfaces only contain abstract methods, there is no need to provide a method body.
B. public final double methoda(); This option is not a valid declaration within an interface definition. The "final" keyword cannot be used to modify a method declaration within an interface. The "final" keyword is used to indicate that a method cannot be overridden by a subclass.
C. static void methoda(double d1); This option is not a valid declaration within an interface definition. The "static" keyword cannot be used to modify a method declaration within an interface. Static methods belong to the class itself, not an instance of the class.
D. protected void methoda(double d1); This option is not a valid declaration within an interface definition. The "protected" keyword cannot be used to modify a method declaration within an interface. Protected methods are accessible within the same package and by subclasses.
Therefore, the valid declaration within an interface definition is:
The Answer is: A. public double methoda();
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boolean b1 = 0;
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boolean b2 = 'false';
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boolean b3 = false;
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boolean b4 = Boolean.false();
A boolean can only be assigned the literal true or false.
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boolean b1 = 0;
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boolean b2 = 'false';
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boolean b3 = false;
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boolean b4 = Boolean.false();
A boolean can only be assigned the literal true or false.
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boolean b1 = 0;
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boolean b2 = 'false';
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boolean b3 = false;
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boolean b4 = Boolean.false();
A boolean can only be assigned the literal true or false.
Which is a valid declarations of a String?
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String s1 = null;
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String s2 = 'null';
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String s3 = (String) 'abc';
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String s4 = (String) '\ufeed';
To solve this question, the user needs to understand the syntax and rules for declaring a String in Java.
Let's go through each option and explain why it is right or wrong:
A. String s1 = null; This option is a valid declaration of a String. In Java, a String variable can be assigned a null value, which means it does not reference any object. This option is correct.
B. String s2 = 'null'; This option is incorrect. In Java, single quotes (' ') are used to represent characters, not Strings. To declare a String, double quotes (" ") should be used. Therefore, this option is wrong.
C. String s3 = (String) 'abc'; This option is incorrect. In Java, to declare a String, the value must be enclosed in double quotes (" "). The syntax (String) is used for type casting, which is not necessary when declaring a String. Therefore, this option is wrong.
D. String s4 = (String) '\ufeed'; This option is incorrect. The value '\ufeed' is an escape sequence representing a Unicode character. However, to declare a String, the value must be enclosed in double quotes (" "). Therefore, this option is wrong.
The Answer is: A. String s1 = null;
Which is a valid declarations of a String?
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String s1 = null;
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String s2 = 'null';
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String s3 = (String) 'abc';
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String s4 = (String) '\ufeed';
To solve this question, the user needs to understand the syntax and rules for declaring a String in Java.
Let's go through each option and explain why it is right or wrong:
A. String s1 = null; This option is a valid declaration of a String. In Java, a String variable can be assigned a null value, which means it does not reference any object. This option is correct.
B. String s2 = 'null'; This option is incorrect. In Java, single quotes (' ') are used to represent characters, not Strings. To declare a String, double quotes (" ") should be used. Therefore, this option is wrong.
C. String s3 = (String) 'abc'; This option is incorrect. In Java, to declare a String, the value must be enclosed in double quotes (" "). The syntax (String) is used for type casting, which is not necessary when declaring a String. Therefore, this option is wrong.
D. String s4 = (String) '\ufeed'; This option is incorrect. The value '\ufeed' is an escape sequence representing a Unicode character. However, to declare a String, the value must be enclosed in double quotes (" "). Therefore, this option is wrong.
The Answer is: A. String s1 = null;
Which is a valid declarations of a String?
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String s1 = null;
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String s2 = 'null';
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String s3 = (String) 'abc';
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String s4 = (String) '\ufeed';
To solve this question, the user needs to understand the syntax and rules for declaring a String in Java.
Let's go through each option and explain why it is right or wrong:
A. String s1 = null; This option is a valid declaration of a String. In Java, a String variable can be assigned a null value, which means it does not reference any object. This option is correct.
B. String s2 = 'null'; This option is incorrect. In Java, single quotes (' ') are used to represent characters, not Strings. To declare a String, double quotes (" ") should be used. Therefore, this option is wrong.
C. String s3 = (String) 'abc'; This option is incorrect. In Java, to declare a String, the value must be enclosed in double quotes (" "). The syntax (String) is used for type casting, which is not necessary when declaring a String. Therefore, this option is wrong.
D. String s4 = (String) '\ufeed'; This option is incorrect. The value '\ufeed' is an escape sequence representing a Unicode character. However, to declare a String, the value must be enclosed in double quotes (" "). Therefore, this option is wrong.
The Answer is: A. String s1 = null;
Which is a valid declarations of a String?
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String s1 = null;
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String s2 = 'null';
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String s3 = (String) 'abc';
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String s4 = (String) '\ufeed';
To solve this question, the user needs to understand the syntax and rules for declaring a String in Java.
Let's go through each option and explain why it is right or wrong:
A. String s1 = null; This option is a valid declaration of a String. In Java, a String variable can be assigned a null value, which means it does not reference any object. This option is correct.
B. String s2 = 'null'; This option is incorrect. In Java, single quotes (' ') are used to represent characters, not Strings. To declare a String, double quotes (" ") should be used. Therefore, this option is wrong.
C. String s3 = (String) 'abc'; This option is incorrect. In Java, to declare a String, the value must be enclosed in double quotes (" "). The syntax (String) is used for type casting, which is not necessary when declaring a String. Therefore, this option is wrong.
D. String s4 = (String) '\ufeed'; This option is incorrect. The value '\ufeed' is an escape sequence representing a Unicode character. However, to declare a String, the value must be enclosed in double quotes (" "). Therefore, this option is wrong.
The Answer is: A. String s1 = null;
Which is a valid declarations of a String?
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String s1 = null;
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String s2 = 'null';
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String s3 = (String) 'abc';
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String s4 = (String) '\ufeed';
To solve this question, the user needs to understand the syntax and rules for declaring a String in Java.
Let's go through each option and explain why it is right or wrong:
A. String s1 = null; This option is a valid declaration of a String. In Java, a String variable can be assigned a null value, which means it does not reference any object. This option is correct.
B. String s2 = 'null'; This option is incorrect. In Java, single quotes (' ') are used to represent characters, not Strings. To declare a String, double quotes (" ") should be used. Therefore, this option is wrong.
C. String s3 = (String) 'abc'; This option is incorrect. In Java, to declare a String, the value must be enclosed in double quotes (" "). The syntax (String) is used for type casting, which is not necessary when declaring a String. Therefore, this option is wrong.
D. String s4 = (String) '\ufeed'; This option is incorrect. The value '\ufeed' is an escape sequence representing a Unicode character. However, to declare a String, the value must be enclosed in double quotes (" "). Therefore, this option is wrong.
The Answer is: A. String s1 = null;
What is the numerical range of a char?
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-128 to 127
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-(215) to (215) - 1
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0 to 32767
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0 to 65535
AI Explanation
To answer this question, we need to understand the numerical range of a char data type in programming.
In most programming languages, including Java and C++, a char data type represents a single character and is stored as a 16-bit value.
The correct answer is D) 0 to 65535. This is because a char data type can store values between 0 and 65535, inclusive. The range starts from 0 because it represents the Unicode value for the character '0', and it goes up to 65535 because it represents the maximum Unicode value that can be stored in a char.
Option A) -128 to 127 is incorrect because this range corresponds to the numerical range of a byte data type, which is 8 bits in size.
Option B) -(2^15) to (2^15) - 1 is incorrect because this range corresponds to the numerical range of a short data type, which is 16 bits in size.
Option C) 0 to 32767 is incorrect because this range corresponds to the numerical range of an int data type, which is typically 32 bits in size.
Therefore, the correct answer is D) 0 to 65535.
What is the numerical range of a char?
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-128 to 127
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-(215) to (215) - 1
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0 to 32767
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0 to 65535
AI Explanation
To answer this question, we need to understand the numerical range of a char data type in programming.
In most programming languages, including Java and C++, a char data type represents a single character and is stored as a 16-bit value.
The correct answer is D) 0 to 65535. This is because a char data type can store values between 0 and 65535, inclusive. The range starts from 0 because it represents the Unicode value for the character '0', and it goes up to 65535 because it represents the maximum Unicode value that can be stored in a char.
Option A) -128 to 127 is incorrect because this range corresponds to the numerical range of a byte data type, which is 8 bits in size.
Option B) -(2^15) to (2^15) - 1 is incorrect because this range corresponds to the numerical range of a short data type, which is 16 bits in size.
Option C) 0 to 32767 is incorrect because this range corresponds to the numerical range of an int data type, which is typically 32 bits in size.
Therefore, the correct answer is D) 0 to 65535.
What is the numerical range of a char?
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-128 to 127
-
-(215) to (215) - 1
-
0 to 32767
-
0 to 65535
AI Explanation
To answer this question, we need to understand the numerical range of a char data type in programming.
In most programming languages, including Java and C++, a char data type represents a single character and is stored as a 16-bit value.
The correct answer is D) 0 to 65535. This is because a char data type can store values between 0 and 65535, inclusive. The range starts from 0 because it represents the Unicode value for the character '0', and it goes up to 65535 because it represents the maximum Unicode value that can be stored in a char.
Option A) -128 to 127 is incorrect because this range corresponds to the numerical range of a byte data type, which is 8 bits in size.
Option B) -(2^15) to (2^15) - 1 is incorrect because this range corresponds to the numerical range of a short data type, which is 16 bits in size.
Option C) 0 to 32767 is incorrect because this range corresponds to the numerical range of an int data type, which is typically 32 bits in size.
Therefore, the correct answer is D) 0 to 65535.
What is the numerical range of a char?
-
-128 to 127
-
-(215) to (215) - 1
-
0 to 32767
-
0 to 65535
AI Explanation
To answer this question, we need to understand the numerical range of a char data type in programming.
In most programming languages, including Java and C++, a char data type represents a single character and is stored as a 16-bit value.
The correct answer is D) 0 to 65535. This is because a char data type can store values between 0 and 65535, inclusive. The range starts from 0 because it represents the Unicode value for the character '0', and it goes up to 65535 because it represents the maximum Unicode value that can be stored in a char.
Option A) -128 to 127 is incorrect because this range corresponds to the numerical range of a byte data type, which is 8 bits in size.
Option B) -(2^15) to (2^15) - 1 is incorrect because this range corresponds to the numerical range of a short data type, which is 16 bits in size.
Option C) 0 to 32767 is incorrect because this range corresponds to the numerical range of an int data type, which is typically 32 bits in size.
Therefore, the correct answer is D) 0 to 65535.
You want subclasses in any package to have access to members of a superclass. Which is the most restrictive access that accomplishes this objective?
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public
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protected
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private
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transient
protected makes a member accessible only to classes in the same package or subclass of the class
You want subclasses in any package to have access to members of a superclass. Which is the most restrictive access that accomplishes this objective?
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public
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protected
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private
-
transient
protected makes a member accessible only to classes in the same package or subclass of the class
What is the most restrictive access modifier that will allow members of one class to have access to members of another class in the same package?
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public
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abstract
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protected
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synchronized
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default access
default access is the "package oriented" access modifier.
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int[ ] scores = {3, 5, 7};
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int [ ][ ] scores = {2,7,6}, {9,3,45};
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String cats[ ] = {"Fluffy", "Spot", "Zeus"};
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boolean results[ ] = new boolean [] {true, false, true};
To determine which option will cause a compiler error, we need to understand the syntax and rules of arrays in programming languages.
An array is a data structure that allows you to store multiple values of the same type under a single variable name. In most programming languages, arrays have a fixed size and can only store elements of the same type.
Now, let's analyze each option to identify if it will cause a compiler error:
A. int[ ] scores = {3, 5, 7}; This option is correct. It declares an integer array named "scores" and initializes it with three values: 3, 5, and 7. This code will not cause a compiler error.
B. int [ ][ ] scores = {2,7,6}, {9,3,45}; This option is incorrect. It declares a two-dimensional integer array named "scores" and initializes it with two rows and three columns. However, the array initializer is incorrect. Each row of a two-dimensional array should be enclosed in curly braces. The correct syntax should be: int [ ][ ] scores = {{2,7,6}, {9,3,45}}; The given code will cause a compiler error.
C. String cats[ ] = {"Fluffy", "Spot", "Zeus"}; This option is correct. It declares a String array named "cats" and initializes it with three strings: "Fluffy", "Spot", and "Zeus". This code will not cause a compiler error.
D. boolean results[ ] = new boolean [] {true, false, true}; This option is correct. It declares a boolean array named "results" and initializes it with three boolean values: true, false, and true. This code will not cause a compiler error.
Therefore, the option that will cause a compiler error is:
The Answer is: B
You want a class to have access to members of another class in the same package. Which is the most restrictive access that accomplishes this objective?
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public
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private
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protected
-
default access
- Public access is the least restrictive access modifier. Members with public access can be accessed from anywhere in the program.
- Private access is the most restrictive access modifier. Members with private access can only be accessed from within the class in which they are declared.
- Protected access is a middle ground between public and private access. Members with protected access can be accessed from within the class in which they are declared, and from subclasses of that class.
- Default access is also known as package-private access. Members with default access can be accessed from any class in the same package.
So, the most restrictive access modifier that allows a class in the same package to access members of another class in the same package is Default.
Here is a table that summarizes the different access modifiers in Java:
| Access modifier | Visibility |
|---|---|
| public | Anywhere in the program |
| private | Only within the class in which it is declared |
| protected | Within the class in which it is declared, and from subclasses of that class |
| default | Only within the same package |
The difference between protected and default access is that protected allows access from subclasses in other packages, while default does not. For example, if you have a class Animal in package A, and a class Dog in package B that extends Animal, then the class Dog can access the protected members of Animal, but not the default members. However, if you have a class Cat in package A that extends Animal, then the class Cat can access both the protected and the default members of Animal.
Which one creates an instance of an array?
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int[ ] ia = new int[15];
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float fa = new float[20];
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char[ ] ca = "Some String";
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int ia[ ] [ ] = { 4, 5, 6 }, { 1,2,3 };
Which of the following class level (nonlocal) variable declarations will not compile?
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protected int a;
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transient int b = 3;
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private synchronized int e;
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volatile int d;
AI Explanation
To answer this question, we need to understand the rules and restrictions for declaring class-level (nonlocal) variables in Java.
Let's go through each option to understand why it is correct or incorrect:
Option A) protected int a; This option is correct because there are no syntax errors or restrictions with declaring a protected class-level variable in Java.
Option B) transient int b = 3; This option is correct because there are no syntax errors or restrictions with declaring a transient class-level variable in Java.
Option C) private synchronized int e; This option is incorrect because it violates the rules for declaring class-level variables. In Java, the synchronized modifier is not allowed for class-level variables. It can only be used for instance-level methods or blocks.
Option D) volatile int d; This option is correct because there are no syntax errors or restrictions with declaring a volatile class-level variable in Java.
Therefore, the correct answer is C) private synchronized int e. This option will not compile because the synchronized modifier is not allowed for class-level variables in Java.
Please select the most appropriate option.
Given a method in a protected class, what access modifier do you use to restrict access to that method to only the other members of the same class?
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final
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static
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private
-
protected
The private access modifier limits access to members of the same class.
Which is a valid declaration within an interface?
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public static short stop = 23;
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protected short stop = 23;
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transient short stop = 23;
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final void madness(short stop);
AI Explanation
To answer this question, you need to understand the rules for declaring variables and methods within an interface. Let's go through each option to understand why it is correct or incorrect:
Option A) public static short stop = 23; - This option is correct. In an interface, you can declare variables with the "public static" modifiers. The variable "stop" is declared as a public static short and assigned the value 23.
Option B) protected short stop = 23; - This option is incorrect. In an interface, variables cannot have access modifiers like "protected" or "private". They are implicitly public and static.
Option C) transient short stop = 23; - This option is incorrect. In an interface, variables cannot have modifiers like "transient". They can only have "public" and "static" modifiers.
Option D) final void madness(short stop); - This option is incorrect. In an interface, you can only declare method signatures, not method implementations. The "void" return type and the presence of the method body make this option invalid.
The correct answer is Option A) public static short stop = 23;. This option is correct because it follows the rules for declaring variables within an interface.
Therefore, the correct answer is A) public static short stop = 23;
Suppose that you would like to create an instance of a new Map that has an iteration order that is the same as the iteration order of an existing instance of a Map. Which concrete implementation of the Map interface should be used for the new instance?
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TreeMap
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HashMap
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LinkedHashMap
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The answer depends on the implementation of the existing instance.
To solve this question, the user needs to know the differences between the concrete implementations of the Map interface and their iteration orders.
Now, let's go through each option and explain why it is right or wrong:
A. TreeMap: This option is incorrect because TreeMap sorts its elements in natural order or by a custom Comparator, whereas the iteration order of the existing instance of the Map may not be sorted. Thus, creating a new instance of TreeMap would not preserve the same iteration order as the existing instance.
B. HashMap: This option is incorrect because HashMap does not guarantee any particular iteration order. Thus, creating a new instance of HashMap would not preserve the same iteration order as the existing instance.
C. LinkedHashMap: This option is correct. LinkedHashMap is similar to HashMap but maintains a doubly-linked list running through all its entries, which allows it to maintain the insertion order. Thus, creating a new instance of LinkedHashMap would preserve the same iteration order as the existing instance.
D. The answer depends on the implementation of the existing instance: This option is incorrect because the iteration order of the existing instance of the Map may not be guaranteed by all implementations of the Map interface. Thus, it is not a reliable solution to assume that any implementation of the Map interface would preserve the same iteration order as the existing instance.
The Answer is: C. LinkedHashMap
Which class does not override the equals() and hashCode() methods, inheriting them directly from class Object?
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java.lang.String
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java.lang.Double
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java.lang.StringBuffer
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java.lang.Character