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All questions of Methods in Java for EmSAT Achieve Exam
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int result = multiply(3, 4);
System.out.println(result);
} public static int multiply(int a, int b) {
if (b == 0) {
return 0;
} else {
return a + multiply(a, b - 1);
}
}
}- a)0
- b)3
- c)4
- d)12
Correct answer is option 'D'. Can you explain this answer?
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int result = multiply(3, 4);
System.out.println(result);
}
public class MyClass {
public static void main(String[] args) {
int result = multiply(3, 4);
System.out.println(result);
}
public static int multiply(int a, int b) {
if (b == 0) {
return 0;
} else {
return a + multiply(a, b - 1);
}
}
}
if (b == 0) {
return 0;
} else {
return a + multiply(a, b - 1);
}
}
}
a)
0
b)
3
c)
4
d)
12
|
Khalil Al Matrooshi answered |
The given code is written in Java and it calculates the multiplication of two integers using a recursive function.
Let's break down the code and understand how it works:
1. The main method is the entry point of the code. It calls the multiply method with arguments 3 and 4 and stores the result in the variable 'result'.
2. The multiply method takes two integers, 'a' and 'b', as parameters. It is a recursive function that calculates the multiplication of 'a' and 'b' by reducing 'b' by 1 in each recursive call until 'b' becomes 0.
3. Inside the multiply method, it checks if 'b' is equal to 0. If it is, it means the multiplication is completed, so it returns 0 as the base case.
4. If 'b' is not equal to 0, it enters the else block. It recursively calls the multiply method with the same value of 'a' and 'b-1'. The return value of this recursive call is multiplied with 'a' and returned.
5. In each recursive call, 'b' is reduced by 1 until it becomes 0, at which point the base case is triggered, and the recursion stops.
6. Finally, the main method prints the value of 'result', which is the multiplication of 3 and 4.
Based on the above explanation, the output of the code will be 12.
Let's break down the code and understand how it works:
1. The main method is the entry point of the code. It calls the multiply method with arguments 3 and 4 and stores the result in the variable 'result'.
2. The multiply method takes two integers, 'a' and 'b', as parameters. It is a recursive function that calculates the multiplication of 'a' and 'b' by reducing 'b' by 1 in each recursive call until 'b' becomes 0.
3. Inside the multiply method, it checks if 'b' is equal to 0. If it is, it means the multiplication is completed, so it returns 0 as the base case.
4. If 'b' is not equal to 0, it enters the else block. It recursively calls the multiply method with the same value of 'a' and 'b-1'. The return value of this recursive call is multiplied with 'a' and returned.
5. In each recursive call, 'b' is reduced by 1 until it becomes 0, at which point the base case is triggered, and the recursion stops.
6. Finally, the main method prints the value of 'result', which is the multiplication of 3 and 4.
Based on the above explanation, the output of the code will be 12.
What is the output of the following code snippet?
public class Example {
public static void main(String[] args) {
int result = sum(3, 4);
System.out.println(result);
} public static int sum(int a, int b) {
return a + b;
} public static double sum(double a, double b) {
return a + b;
}
}- a)7
- b)7.0
- c)Compilation error
- d)Runtime error
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code snippet?
public class Example {
public static void main(String[] args) {
int result = sum(3, 4);
System.out.println(result);
}
public class Example {
public static void main(String[] args) {
int result = sum(3, 4);
System.out.println(result);
}
public static int sum(int a, int b) {
return a + b;
}
return a + b;
}
public static double sum(double a, double b) {
return a + b;
}
}
return a + b;
}
}
a)
7
b)
7.0
c)
Compilation error
d)
Runtime error
|
Sonal Yadav answered |
The 'sum' method in the code snippet is overloaded with two versions: one that takes two integers and another that takes two doubles. In the 'main' method, 'sum(3, 4)' is called, and since the arguments are integers, the first version of 'sum' is invoked, returning the sum of 3 and 4, which is 7.
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
printMessage("Hello");
printMessage("Hi", 3);
} public static void printMessage(String message) {
System.out.println(message);
} public static void printMessage(String message, int count) {
for (int i = 0; i < count; i++) {
System.out.println(message);
}
}
}- a)Hello
Hi - b)Hello
Hi Hi Hi - c)Compilation Error
- d)Runtime Error
Correct answer is option 'B'. Can you explain this answer?
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
printMessage("Hello");
printMessage("Hi", 3);
}
public class MyClass {
public static void main(String[] args) {
printMessage("Hello");
printMessage("Hi", 3);
}
public static void printMessage(String message) {
System.out.println(message);
}
System.out.println(message);
}
public static void printMessage(String message, int count) {
for (int i = 0; i < count; i++) {
System.out.println(message);
}
}
}
for (int i = 0; i < count; i++) {
System.out.println(message);
}
}
}
a)
Hello
Hi
Hi
b)
Hello
Hi Hi Hi
Hi Hi Hi
c)
Compilation Error
d)
Runtime Error
|
|
Sonal Yadav answered |
The 'printMessage' method is overloaded to accept either a single string or a string and an integer. In the 'main' method, the first call to 'printMessage' will print "Hello," and the second call will print "Hi" three times.
Which of the following is true about passing arguments by value in Java?- a)The changes made to the passed argument within the method are reflected outside the method.
- b)Only primitive data types can be passed by value in Java.
- c)Objects are always passed by value in Java.
- d)Passing arguments by value means passing the actual memory address of the variable.
Correct answer is option 'C'. Can you explain this answer?
Which of the following is true about passing arguments by value in Java?
a)
The changes made to the passed argument within the method are reflected outside the method.
b)
Only primitive data types can be passed by value in Java.
c)
Objects are always passed by value in Java.
d)
Passing arguments by value means passing the actual memory address of the variable.
|
Amna Al Jabri answered |
Understanding Passing Arguments by Value in Java
In Java, all arguments passed to methods are handled by value. This concept can sometimes be misunderstood, especially in the context of objects. Here’s a detailed explanation of why option 'C' is correct.
What Does "Passed by Value" Mean?
- Definition: "Passed by value" refers to the fact that a copy of the variable’s value is made when passed to a method.
- Primitive Data Types: For primitive types (int, char, etc.), the actual value is copied.
- Objects: For objects, the reference (memory address) to the object is copied, not the object itself.
Why Option 'C' is Correct
- Objects Are Passed by Value: In Java, when an object is passed to a method, a copy of the reference to that object is made. Therefore, changes made to the object's attributes within the method will affect the original object because both the original and the copied reference point to the same object in memory.
- Example: If you modify an object’s field inside a method, the changes will persist outside the method. However, if you reassign the reference to point to a new object, this change will not affect the original reference.
Why Other Options Are Incorrect
- Option A: Changes made to the argument within the method are reflected only if you modify the object itself, not if you reassign the reference.
- Option B: Both primitive data types and object references are passed by value in Java.
- Option D: Passing by value does not mean passing the actual memory address; it means passing a copy of the reference.
Understanding these distinctions helps clarify how Java handles method arguments and object manipulation.
In Java, all arguments passed to methods are handled by value. This concept can sometimes be misunderstood, especially in the context of objects. Here’s a detailed explanation of why option 'C' is correct.
What Does "Passed by Value" Mean?
- Definition: "Passed by value" refers to the fact that a copy of the variable’s value is made when passed to a method.
- Primitive Data Types: For primitive types (int, char, etc.), the actual value is copied.
- Objects: For objects, the reference (memory address) to the object is copied, not the object itself.
Why Option 'C' is Correct
- Objects Are Passed by Value: In Java, when an object is passed to a method, a copy of the reference to that object is made. Therefore, changes made to the object's attributes within the method will affect the original object because both the original and the copied reference point to the same object in memory.
- Example: If you modify an object’s field inside a method, the changes will persist outside the method. However, if you reassign the reference to point to a new object, this change will not affect the original reference.
Why Other Options Are Incorrect
- Option A: Changes made to the argument within the method are reflected only if you modify the object itself, not if you reassign the reference.
- Option B: Both primitive data types and object references are passed by value in Java.
- Option D: Passing by value does not mean passing the actual memory address; it means passing a copy of the reference.
Understanding these distinctions helps clarify how Java handles method arguments and object manipulation.
What is the difference between method overloading and method overriding in Java?- a)Method overloading occurs within the same class, while method overriding occurs between a superclass and its subclass.
- b)Method overloading is done by redefining a method with the same name but different parameter lists, while method overriding is done by redefining a method in a subclass with the same signature as a method in its superclass.
- c)Method overloading allows polymorphism, while method overriding does not.
- d)Method overloading is used for static methods, while method overriding is used for instance methods.
Correct answer is option 'B'. Can you explain this answer?
What is the difference between method overloading and method overriding in Java?
a)
Method overloading occurs within the same class, while method overriding occurs between a superclass and its subclass.
b)
Method overloading is done by redefining a method with the same name but different parameter lists, while method overriding is done by redefining a method in a subclass with the same signature as a method in its superclass.
c)
Method overloading allows polymorphism, while method overriding does not.
d)
Method overloading is used for static methods, while method overriding is used for instance methods.
|
|
Saeed Al Farsi answered |
Method Overloading:
Method overloading is a feature in Java that allows a class to have multiple methods with the same name but different parameters. These methods can perform similar tasks but with different sets of inputs. The compiler decides which method to call based on the number and types of arguments passed to it.
Method Overriding:
Method overriding is a feature in Java that allows a subclass to provide a different implementation of a method that is already defined in its superclass. The method in the subclass must have the same name, return type, and parameters as the method in the superclass. The purpose of method overriding is to provide a specific implementation of the method in the subclass that is different from the one in the superclass.
Differences between Method Overloading and Method Overriding:
1. Scope:
- Method overloading occurs within the same class, where multiple methods with the same name but different parameters are defined.
- Method overriding occurs between a superclass and its subclass, where the subclass provides a different implementation of a method that is already defined in the superclass.
2. Definition:
- Method overloading is done by redefining a method with the same name but different parameter lists.
- Method overriding is done by redefining a method in a subclass with the same signature as a method in its superclass.
3. Polymorphism:
- Method overloading allows polymorphism, as the compiler determines which method to call based on the arguments passed.
- Method overriding does not allow polymorphism, as the method to be called is determined at runtime based on the actual object type.
4. Usage:
- Method overloading is used for both static and instance methods.
- Method overriding is only used for instance methods.
In conclusion, method overloading and method overriding are both important features in Java, but they have different purposes and usage scenarios. Method overloading is used to provide multiple methods with the same name but different parameters within the same class, while method overriding is used to redefine a method in a subclass with the same signature as a method in its superclass.
Method overloading is a feature in Java that allows a class to have multiple methods with the same name but different parameters. These methods can perform similar tasks but with different sets of inputs. The compiler decides which method to call based on the number and types of arguments passed to it.
Method Overriding:
Method overriding is a feature in Java that allows a subclass to provide a different implementation of a method that is already defined in its superclass. The method in the subclass must have the same name, return type, and parameters as the method in the superclass. The purpose of method overriding is to provide a specific implementation of the method in the subclass that is different from the one in the superclass.
Differences between Method Overloading and Method Overriding:
1. Scope:
- Method overloading occurs within the same class, where multiple methods with the same name but different parameters are defined.
- Method overriding occurs between a superclass and its subclass, where the subclass provides a different implementation of a method that is already defined in the superclass.
2. Definition:
- Method overloading is done by redefining a method with the same name but different parameter lists.
- Method overriding is done by redefining a method in a subclass with the same signature as a method in its superclass.
3. Polymorphism:
- Method overloading allows polymorphism, as the compiler determines which method to call based on the arguments passed.
- Method overriding does not allow polymorphism, as the method to be called is determined at runtime based on the actual object type.
4. Usage:
- Method overloading is used for both static and instance methods.
- Method overriding is only used for instance methods.
In conclusion, method overloading and method overriding are both important features in Java, but they have different purposes and usage scenarios. Method overloading is used to provide multiple methods with the same name but different parameters within the same class, while method overriding is used to redefine a method in a subclass with the same signature as a method in its superclass.
Which of the following is a valid method declaration in Java?- a)'void myMethod() {}'
- b)'int myMethod() {}'
- c)'myMethod() {}'
- d)'myMethod(void) {}'
Correct answer is option 'A'. Can you explain this answer?
Which of the following is a valid method declaration in Java?
a)
'void myMethod() {}'
b)
'int myMethod() {}'
c)
'myMethod() {}'
d)
'myMethod(void) {}'
|
Mubarak Al Khalifa answered |
Valid Method Declaration in Java
A method in Java is a block of code that performs a specific task. It is declared inside a class and can be called or invoked whenever required. A method can have a return type, parameters, and an optional access specifier.
Among the given options (a, b, c, and d), only option 'a' is a valid method declaration in Java. Let's understand why other options are not valid:
a) void myMethod() {}
- This is a valid method declaration in Java.
- The method name is "myMethod" and it has no parameters.
- The return type is "void", which means the method does not return any value.
b) int myMethod() {}
- This is not a valid method declaration.
- The return type is "int", which means the method is expected to return an integer value.
- However, the method body is empty, and there is no return statement. This results in a compilation error.
c) myMethod() {}
- This is not a valid method declaration.
- The return type is missing, which is a mandatory part of a method declaration.
- It should specify the type of value the method will return or use "void" if it does not return any value.
- Without a return type, the declaration is incomplete and will cause a compilation error.
d) myMethod(void) {}
- This is not a valid method declaration.
- The parameter declaration is incorrect.
- In Java, the parameter type should be specified before the parameter name.
- The correct syntax would be "myMethod(void parameterName)" or "myMethod(parameterType parameterName)".
- Additionally, using "void" as a parameter type is not allowed in Java.
Conclusion
In Java, a valid method declaration should include the return type (or "void" if no return value), the method name, and optionally the parameters. Among the given options, only option 'a' satisfies these requirements and is a valid method declaration.
A method in Java is a block of code that performs a specific task. It is declared inside a class and can be called or invoked whenever required. A method can have a return type, parameters, and an optional access specifier.
Among the given options (a, b, c, and d), only option 'a' is a valid method declaration in Java. Let's understand why other options are not valid:
a) void myMethod() {}
- This is a valid method declaration in Java.
- The method name is "myMethod" and it has no parameters.
- The return type is "void", which means the method does not return any value.
b) int myMethod() {}
- This is not a valid method declaration.
- The return type is "int", which means the method is expected to return an integer value.
- However, the method body is empty, and there is no return statement. This results in a compilation error.
c) myMethod() {}
- This is not a valid method declaration.
- The return type is missing, which is a mandatory part of a method declaration.
- It should specify the type of value the method will return or use "void" if it does not return any value.
- Without a return type, the declaration is incomplete and will cause a compilation error.
d) myMethod(void) {}
- This is not a valid method declaration.
- The parameter declaration is incorrect.
- In Java, the parameter type should be specified before the parameter name.
- The correct syntax would be "myMethod(void parameterName)" or "myMethod(parameterType parameterName)".
- Additionally, using "void" as a parameter type is not allowed in Java.
Conclusion
In Java, a valid method declaration should include the return type (or "void" if no return value), the method name, and optionally the parameters. Among the given options, only option 'a' satisfies these requirements and is a valid method declaration.
Which of the following is true about method parameters in Java?- a)Parameters are optional in method declarations.
- b)A method can have multiple parameters of the same name.
- c)Parameters can only be of primitive data types.
- d)Parameters are used to specify the return type of a method.
Correct answer is option 'B'. Can you explain this answer?
Which of the following is true about method parameters in Java?
a)
Parameters are optional in method declarations.
b)
A method can have multiple parameters of the same name.
c)
Parameters can only be of primitive data types.
d)
Parameters are used to specify the return type of a method.
|
|
Sonal Yadav answered |
A method in Java can have multiple parameters, and they can have the same name. The parameters act as variables within the method's scope.
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int x = 5;
int y = increment(x);
System.out.println(y);
} public static int increment(int value) {
return ++value;
}
}- a)5
- b)6
- c)Compilation Error
- d)Runtime Error
Correct answer is option 'B'. Can you explain this answer?
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int x = 5;
int y = increment(x);
System.out.println(y);
}
public class MyClass {
public static void main(String[] args) {
int x = 5;
int y = increment(x);
System.out.println(y);
}
public static int increment(int value) {
return ++value;
}
}
return ++value;
}
}
a)
5
b)
6
c)
Compilation Error
d)
Runtime Error
|
|
Khalid Al Khouri answered |
Explanation:
The given code defines a class named "MyClass" with a main method.
The main method initializes an integer variable "x" with the value 5.
Then, it calls the increment method and passes the value of "x" as an argument.
The returned value from the increment method is stored in the variable "y".
Finally, the value of "y" is printed.
Step-by-Step Execution:
1. The main method is called and the variable "x" is initialized with the value 5.
2. The increment method is called, and the value of "x" (i.e., 5) is passed as an argument.
3. In the increment method, the value parameter is received and stored in the variable "value".
4. The value of "value" is returned.
5. The returned value (i.e., 5) is stored in the variable "y".
6. The value of "y" (i.e., 5) is printed using the System.out.println statement.
Output:
The output of the code will be:
6
Explanation:
The output is 6 because the increment method does not actually increment the value passed to it. It simply returns the value as it is. So, when the value 5 is passed to the increment method, it returns 5, and that value is stored in the variable "y". Therefore, the output is 6.
The given code defines a class named "MyClass" with a main method.
The main method initializes an integer variable "x" with the value 5.
Then, it calls the increment method and passes the value of "x" as an argument.
The returned value from the increment method is stored in the variable "y".
Finally, the value of "y" is printed.
Step-by-Step Execution:
1. The main method is called and the variable "x" is initialized with the value 5.
2. The increment method is called, and the value of "x" (i.e., 5) is passed as an argument.
3. In the increment method, the value parameter is received and stored in the variable "value".
4. The value of "value" is returned.
5. The returned value (i.e., 5) is stored in the variable "y".
6. The value of "y" (i.e., 5) is printed using the System.out.println statement.
Output:
The output of the code will be:
6
Explanation:
The output is 6 because the increment method does not actually increment the value passed to it. It simply returns the value as it is. So, when the value 5 is passed to the increment method, it returns 5, and that value is stored in the variable "y". Therefore, the output is 6.
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int x = 10;
changeValue(x);
System.out.println(x);
} public static void changeValue(int value) {
value = 20;
}
}- a)10
- b)20
- c)Compilation Error
- d)Runtime Error
Correct answer is option 'A'. Can you explain this answer?
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int x = 10;
changeValue(x);
System.out.println(x);
}
public class MyClass {
public static void main(String[] args) {
int x = 10;
changeValue(x);
System.out.println(x);
}
public static void changeValue(int value) {
value = 20;
}
}
value = 20;
}
}
a)
10
b)
20
c)
Compilation Error
d)
Runtime Error
|
Fawaz Al Ameri answered |
Answer:
The output of the code will be 10.
Explanation:
When the code is executed, the following steps occur:
1. The variable x is declared and initialized with the value 10.
2. The method changeValue() is invoked and the value of x is passed as an argument.
3. In the changeValue() method, a new variable value is declared and initialized with the value of x, which is 10.
4. The value of value is then changed to 20.
5. However, this change does not affect the original value of x because Java passes arguments by value, which means that a copy of the variable's value is passed to the method. Therefore, changing the value of the parameter value does not modify the original variable x.
6. The method changeValue() completes execution and the control returns to the main() method.
7. The value of x is printed using the System.out.println() statement, which still retains its original value of 10.
Therefore, the output of the code is 10.
The output of the code will be 10.
Explanation:
When the code is executed, the following steps occur:
1. The variable x is declared and initialized with the value 10.
2. The method changeValue() is invoked and the value of x is passed as an argument.
3. In the changeValue() method, a new variable value is declared and initialized with the value of x, which is 10.
4. The value of value is then changed to 20.
5. However, this change does not affect the original value of x because Java passes arguments by value, which means that a copy of the variable's value is passed to the method. Therefore, changing the value of the parameter value does not modify the original variable x.
6. The method changeValue() completes execution and the control returns to the main() method.
7. The value of x is printed using the System.out.println() statement, which still retains its original value of 10.
Therefore, the output of the code is 10.
Which of the following methods can be used to calculate the factorial of a number in Java?- a)public static int factorial(int n) { ... }
- b)public static long factorial(long n) { ... }
- c)public static int factorial(int n, int m) { ... }
- d)public static int factorial(int[] arr) { ... }
Correct answer is option 'A'. Can you explain this answer?
Which of the following methods can be used to calculate the factorial of a number in Java?
a)
public static int factorial(int n) { ... }
b)
public static long factorial(long n) { ... }
c)
public static int factorial(int n, int m) { ... }
d)
public static int factorial(int[] arr) { ... }
|
Arien Instructors answered |
Both option a) and b) are valid methods to calculate the factorial of a number. The return type of the factorial function can be either 'int' or 'long' depending on the range of values you want to support.
Note: The other options c) and d) are not suitable for calculating the factorial. Option c) includes an extra parameter m that is not required for factorial calculation, and option d) takes an array as a parameter, which is not necessary for the factorial computation.
What is the output of the following code snippet?
public class Example {
public static void main(String[] args) {
int x = 5;
updateValue(x);
System.out.println(x);
} public static void updateValue(int value) {
value = value + 1;
}
}- a)4
- b)5
- c)6
- d)Compilation error
Correct answer is option 'B'. Can you explain this answer?
What is the output of the following code snippet?
public class Example {
public static void main(String[] args) {
int x = 5;
updateValue(x);
System.out.println(x);
}
public class Example {
public static void main(String[] args) {
int x = 5;
updateValue(x);
System.out.println(x);
}
public static void updateValue(int value) {
value = value + 1;
}
}
value = value + 1;
}
}
a)
4
b)
5
c)
6
d)
Compilation error
|
|
Sonal Yadav answered |
The 'updateValue' method in the code snippet takes an integer parameter and increments it by 1. However, the parameter 'value' is a copy of the value of 'x' in the main method. Therefore, modifying 'value' does not affect the original variable 'x'.
Which of the following methods is not valid in Java?- a)public static int add(int a, int
- b){ ... } b) protected void printMessage(String message) { ... }
- c)private static void calculate(double num) { ... }
- d)e) { ... } c) private static void calculate(double num) { ... }
Correct answer is option 'D'. Can you explain this answer?
Which of the following methods is not valid in Java?
a)
public static int add(int a, int
b)
{ ... } b) protected void printMessage(String message) { ... }
c)
private static void calculate(double num) { ... }
d)
e) { ... } c) private static void calculate(double num) { ... }
|
|
Sonal Yadav answered |
The missing access modifier in option d) makes the method invalid. In Java, methods should have an access modifier (e.g., public, private, protected) specified explicitly.
Which of the following statements about methods in Java is correct?- a)A method can only return a single value.
- b)A method can have multiple return statements.
- c)A method can have a void return type and still return a value.
- d)A method cannot be called recursively.
Correct answer is option 'B'. Can you explain this answer?
Which of the following statements about methods in Java is correct?
a)
A method can only return a single value.
b)
A method can have multiple return statements.
c)
A method can have a void return type and still return a value.
d)
A method cannot be called recursively.
|
|
Sonal Yadav answered |
A method in Java can have multiple return statements. The return statement is used to return a value from the method and terminate its execution.
What is the difference between call by value and call by reference in Java?- a)Call by value allows modification of the original variable, while call by reference does not.
- b)Call by reference allows modification of the original variable, while call by value does not.
- c)Call by value passes a copy of the variable, while call by reference passes the actual variable.
- d)Call by reference passes a copy of the variable, while call by value passes the actual variable.
Correct answer is option 'C'. Can you explain this answer?
What is the difference between call by value and call by reference in Java?
a)
Call by value allows modification of the original variable, while call by reference does not.
b)
Call by reference allows modification of the original variable, while call by value does not.
c)
Call by value passes a copy of the variable, while call by reference passes the actual variable.
d)
Call by reference passes a copy of the variable, while call by value passes the actual variable.
|
|
Sonal Yadav answered |
In Java, when a method is called with call by value, a copy of the variable's value is passed to the method. Any modifications made to the parameter within the method do not affect the original variable. Call by reference, on the other hand, passes the actual variable itself, allowing modifications to affect the original variable.
What is the difference between a method declaration and a method definition in Java?- a)A method declaration includes the method's return type, name, and parameter list, while a method definition includes the method's implementation.
- b)A method declaration includes the method's implementation, while a method definition includes the method's return type, name, and parameter list.
- c)There is no difference between a method declaration and a method definition.
- d)A method declaration includes the method's access modifiers, while a method definition does not.
Correct answer is option 'A'. Can you explain this answer?
What is the difference between a method declaration and a method definition in Java?
a)
A method declaration includes the method's return type, name, and parameter list, while a method definition includes the method's implementation.
b)
A method declaration includes the method's implementation, while a method definition includes the method's return type, name, and parameter list.
c)
There is no difference between a method declaration and a method definition.
d)
A method declaration includes the method's access modifiers, while a method definition does not.
|
|
Sonal Yadav answered |
In Java, a method declaration specifies the method's return type, name, and parameter list, while a method definition includes the implementation of the method, including the method body.
What is the return type of a method that does not return any value?- a)void
- b)null
- c)int
- d)No return type is required
Correct answer is option 'A'. Can you explain this answer?
What is the return type of a method that does not return any value?
a)
void
b)
null
c)
int
d)
No return type is required
|
|
Sonal Yadav answered |
When a method does not return any value, its return type is specified as 'void' in Java.
What is the purpose of method parameters in Java?- a)They define the return type of a method.
- b)They provide a way to pass data to a method.
- c)They determine the access level of a method.
- d)They specify the name of a method.
Correct answer is option 'B'. Can you explain this answer?
What is the purpose of method parameters in Java?
a)
They define the return type of a method.
b)
They provide a way to pass data to a method.
c)
They determine the access level of a method.
d)
They specify the name of a method.
|
|
Sonal Yadav answered |
Method parameters in Java provide a way to pass data to a method. They act as placeholders for the values that are passed when the method is called.
What is the difference between instance methods and static methods in Java?- a)Instance methods can only be called from static methods.
- b)Static methods can access instance variables.
- c)Static methods are defined with the "static" keyword.
- d)Instance methods are used for object creation.
Correct answer is option 'B'. Can you explain this answer?
What is the difference between instance methods and static methods in Java?
a)
Instance methods can only be called from static methods.
b)
Static methods can access instance variables.
c)
Static methods are defined with the "static" keyword.
d)
Instance methods are used for object creation.
|
|
Sonal Yadav answered |
Instance methods are associated with an instance of a class and can access instance variables. Static methods, on the other hand, are not associated with any specific instance and cannot access instance variables directly.
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
printNumber(5);
} public static void printNumber(int num) {
if (num <= 0) {
return;
}
System.out.println(num);
printNumber(num - 1);
}
}- a)5 4 3 2 1
- b)1 2 3 4 5
- c)5
- d)Compilation Error
Correct answer is option 'A'. Can you explain this answer?
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
printNumber(5);
}
public class MyClass {
public static void main(String[] args) {
printNumber(5);
}
public static void printNumber(int num) {
if (num <= 0) {
return;
}
System.out.println(num);
printNumber(num - 1);
}
}
if (num <= 0) {
return;
}
System.out.println(num);
printNumber(num - 1);
}
}
a)
5 4 3 2 1
b)
1 2 3 4 5
c)
5
d)
Compilation Error
|
|
Sonal Yadav answered |
The 'printNumber' method is a recursive method that prints the number passed as an argument and then recursively calls itself with 'num - 1'. The recursion continues until 'num' becomes 0 or negative.
What is method overriding in Java?- a)Defining multiple methods with the same name but different return types.
- b)Defining multiple methods with the same name but different parameter lists.
- c)Defining multiple methods with the same name but different access modifiers.
- d)Redefining a method in a subclass with the same signature as a method in its superclass.
Correct answer is option 'D'. Can you explain this answer?
What is method overriding in Java?
a)
Defining multiple methods with the same name but different return types.
b)
Defining multiple methods with the same name but different parameter lists.
c)
Defining multiple methods with the same name but different access modifiers.
d)
Redefining a method in a subclass with the same signature as a method in its superclass.
|
|
Sonal Yadav answered |
Method overriding in Java occurs when a subclass provides its own implementation of a method that is already defined in its superclass. The overriding method has the same name, return type, and parameter list as the method in the superclass.
What is the output of the following code?
public class MyClass {
public static void main(String[] args) {
int result = sum(3, 4);
System.out.println(result);
} public static int sum(int a, int b) {
return a + b;
}
}- a)3
- b)4
- c)7
- d)Compilation Error
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code?
public class MyClass {
public static void main(String[] args) {
int result = sum(3, 4);
System.out.println(result);
}
public class MyClass {
public static void main(String[] args) {
int result = sum(3, 4);
System.out.println(result);
}
public static int sum(int a, int b) {
return a + b;
}
}
return a + b;
}
}
a)
3
b)
4
c)
7
d)
Compilation Error
|
|
Sonal Yadav answered |
The 'sum' method takes two integers as parameters and returns their sum. The 'main' method calls the 'sum' method with arguments 3 and 4, so the output will be the sum of these numbers, which is 7.
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int result = calculate(5, 3);
System.out.println(result);
} public static int calculate(int a, int b) {
int result = a - b;
return result;
}
}- a)2
- b)3
- c)5
- d)Compilation Error
Correct answer is option 'A'. Can you explain this answer?
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int result = calculate(5, 3);
System.out.println(result);
}
public class MyClass {
public static void main(String[] args) {
int result = calculate(5, 3);
System.out.println(result);
}
public static int calculate(int a, int b) {
int result = a - b;
return result;
}
}
int result = a - b;
return result;
}
}
a)
2
b)
3
c)
5
d)
Compilation Error
|
|
Sonal Yadav answered |
The 'calculate' method takes two integers as parameters, subtracts the second number from the first, and returns the result. In the 'main' method, the 'calculate' method is called with arguments 5 and 3, resulting in the value 2, which is printed.
What is the output of the following code snippet?
public class Example {
public static void main(String[] args) {
int result = multiply(3, 4);
System.out.println(result);
}
public static int multiply(int a, int b) {
return a * b;
}
}
- a)7
- b)10
- c)24
- d)12
Correct answer is option 'D'. Can you explain this answer?
What is the output of the following code snippet?
public class Example {
public static void main(String[] args) {
int result = multiply(3, 4);
System.out.println(result);
}
public class Example {
public static void main(String[] args) {
int result = multiply(3, 4);
System.out.println(result);
}
public static int multiply(int a, int b) {
return a * b;
}
}
return a * b;
}
}
a)
7
b)
10
c)
24
d)
12
|
|
Sonal Yadav answered |
The 'multiply' method in the code snippet takes two integer parameters and returns their product. In the 'main' 'method, 'multiply(3, 4)' is called, and the returned value, 12, is printed.
Can a method call itself directly or indirectly in Java?- a)Yes, but it can lead to an infinite loop.
- b)Yes, as long as the method has a return type of 'void'.
- c)No, it is not allowed in Java.
- d)Yes, but only if the method is defined as static.
Correct answer is option 'A'. Can you explain this answer?
Can a method call itself directly or indirectly in Java?
a)
Yes, but it can lead to an infinite loop.
b)
Yes, as long as the method has a return type of 'void'.
c)
No, it is not allowed in Java.
d)
Yes, but only if the method is defined as static.
|
|
Sonal Yadav answered |
Recursive methods in Java can call themselves directly or indirectly. However, without proper termination conditions (base cases), recursive calls can lead to an infinite loop and cause a stack overflow error.
What is method overloading in Java?- a)Defining multiple methods with the same name but different return types.
- b)Defining multiple methods with the same name and the same number of parameters.
- c)Defining multiple methods with the same name and different parameters.
- d)Defining multiple methods with the same name and the same parameters, but in different classes.
Correct answer is option 'C'. Can you explain this answer?
What is method overloading in Java?
a)
Defining multiple methods with the same name but different return types.
b)
Defining multiple methods with the same name and the same number of parameters.
c)
Defining multiple methods with the same name and different parameters.
d)
Defining multiple methods with the same name and the same parameters, but in different classes.
|
|
Sonal Yadav answered |
Method overloading in Java involves defining multiple methods with the same name but different parameters. The parameters can differ in terms of number, types, or order.
What is method overloading resolution in Java?- a)It is the process of determining which method to call based on the method's return type.
- b)It is the process of determining which method to call based on the method's access modifier.
- c)It is the process of determining which method to call based on the method's parameter list.
- d)It is the process of determining which method to call based on the method's name.
Correct answer is option 'C'. Can you explain this answer?
What is method overloading resolution in Java?
a)
It is the process of determining which method to call based on the method's return type.
b)
It is the process of determining which method to call based on the method's access modifier.
c)
It is the process of determining which method to call based on the method's parameter list.
d)
It is the process of determining which method to call based on the method's name.
|
|
Sonal Yadav answered |
Method overloading resolution in Java is the process of determining which method to call when multiple methods with the same name but different parameter lists are available. The compiler matches the arguments provided during the method call with the method's parameter types to determine the most appropriate method to invoke.
Can a method in Java have multiple return statements?- a)Yes, but it is considered a bad programming practice.
- b)No, a method can only have a single return statement.
- c)Yes, multiple return statements provide flexibility in implementing conditional logic.
- d)Yes, but it is only allowed for methods with a return type of 'void'.
Correct answer is option 'C'. Can you explain this answer?
Can a method in Java have multiple return statements?
a)
Yes, but it is considered a bad programming practice.
b)
No, a method can only have a single return statement.
c)
Yes, multiple return statements provide flexibility in implementing conditional logic.
d)
Yes, but it is only allowed for methods with a return type of 'void'.
|
|
Sonal Yadav answered |
In Java, a method can have multiple return statements. Multiple return statements allow flexibility in implementing conditional logic, where different values are returned based on different conditions.
What is method hiding in Java?- a)It is a mechanism to prevent access to methods from outside the class.
- b)It is a way to redefine a method in a subclass with the same signature as a method in its superclass.
- c)It is a way to change the behavior of a method in a subclass without modifying its implementation.
- d)It is a way to define a static method in a subclass with the same signature as a static method in its superclass.
Correct answer is option 'D'. Can you explain this answer?
What is method hiding in Java?
a)
It is a mechanism to prevent access to methods from outside the class.
b)
It is a way to redefine a method in a subclass with the same signature as a method in its superclass.
c)
It is a way to change the behavior of a method in a subclass without modifying its implementation.
d)
It is a way to define a static method in a subclass with the same signature as a static method in its superclass.
|
|
Sonal Yadav answered |
Method hiding in Java occurs when a subclass defines a static method with the same signature as a static method in its superclass. The static method in the subclass "hides" the static method in the superclass.
Which of the following statements about methods in Java is correct?- a)Methods can only be defined within a class.
- b)Methods can only be called from within the same class.
- c)Methods can only return primitive data types.
- d)Methods can have multiple return statements.
Correct answer is option 'A'. Can you explain this answer?
Which of the following statements about methods in Java is correct?
a)
Methods can only be defined within a class.
b)
Methods can only be called from within the same class.
c)
Methods can only return primitive data types.
d)
Methods can have multiple return statements.
|
|
Sonal Yadav answered |
In Java, methods are defined within classes and can be called from other classes as well.
Which of the following is true about method signatures in Java?- a)Method signatures include the return type and method name.
- b)Method signatures include the method name and parameter types.
- c)Method signatures include the method name and access modifiers.
- d)Method signatures include the return type, method name, and parameter types.
Correct answer is option 'B'. Can you explain this answer?
Which of the following is true about method signatures in Java?
a)
Method signatures include the return type and method name.
b)
Method signatures include the method name and parameter types.
c)
Method signatures include the method name and access modifiers.
d)
Method signatures include the return type, method name, and parameter types.
|
|
Sonal Yadav answered |
The method signature in Java consists of the method name and the parameter types. It does not include the return type or the access modifiers.
Which of the following is true about recursive methods in Java?- a)Recursive methods cannot call themselves.
- b)Recursive methods must have a return type of 'void'.
- c)Recursive methods must have at least one base case.
- d)Recursive methods can only be used for sorting algorithms.
Correct answer is option 'C'. Can you explain this answer?
Which of the following is true about recursive methods in Java?
a)
Recursive methods cannot call themselves.
b)
Recursive methods must have a return type of 'void'.
c)
Recursive methods must have at least one base case.
d)
Recursive methods can only be used for sorting algorithms.
|
|
Sonal Yadav answered |
Recursive methods in Java must have at least one base case, which is a condition that terminates the recursion. Without a base case, the recursive method would continue calling itself indefinitely, leading to a stack overflow error.
What is the purpose of the "static" keyword in Java method declarations?- a)It allows the method to be overridden by subclasses.
- b)It indicates that the method is accessible only within the class.
- c)It allows the method to be called without creating an instance of the class.
- d)It prevents the method from being overloaded.
Correct answer is option 'C'. Can you explain this answer?
What is the purpose of the "static" keyword in Java method declarations?
a)
It allows the method to be overridden by subclasses.
b)
It indicates that the method is accessible only within the class.
c)
It allows the method to be called without creating an instance of the class.
d)
It prevents the method from being overloaded.
|
|
Sonal Yadav answered |
The "static" keyword in a Java method declaration allows the method to be called without creating an instance of the class. Static methods belong to the class itself rather than individual instances of the class.
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int x = 5;
int y = 10;
swap(x, y);
System.out.println(x + " " + y);
} public static void swap(int a, int b) {
int temp = a;
a = b;
b = temp;
}
}- a)5 10
- b)10 5
- c)Compilation Error
- d)Runtime Error
Correct answer is option 'A'. Can you explain this answer?
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int x = 5;
int y = 10;
swap(x, y);
System.out.println(x + " " + y);
}
public class MyClass {
public static void main(String[] args) {
int x = 5;
int y = 10;
swap(x, y);
System.out.println(x + " " + y);
}
public static void swap(int a, int b) {
int temp = a;
a = b;
b = temp;
}
}
int temp = a;
a = b;
b = temp;
}
}
a)
5 10
b)
10 5
c)
Compilation Error
d)
Runtime Error
|
|
Sonal Yadav answered |
The 'swap' method takes two integers as parameters and swaps their values using a temporary variable. In the 'main' method, the variables 'x' and 'y' are passed to the 'swap' method. However, the method operates on copies of the variables, so the original values of 'x' and 'y' remain unchanged.
What is the purpose of the "return" keyword in a method?- a)It specifies the access level of the method.
- b)It terminates the execution of the method.
- c)It returns a value from the method.
- d)It defines the parameters of the method.
Correct answer is option 'C'. Can you explain this answer?
What is the purpose of the "return" keyword in a method?
a)
It specifies the access level of the method.
b)
It terminates the execution of the method.
c)
It returns a value from the method.
d)
It defines the parameters of the method.
|
|
Sonal Yadav answered |
The "return" keyword is used to return a value from a method. It terminates the method's execution and sends the specified value back to the caller.
What is the scope of a local variable declared inside a method?- a)It is accessible only within the method in which it is declared.
- b)It is accessible within the entire class.
- c)It is accessible within the same package.
- d)It is accessible throughout the program.
Correct answer is option 'A'. Can you explain this answer?
What is the scope of a local variable declared inside a method?
a)
It is accessible only within the method in which it is declared.
b)
It is accessible within the entire class.
c)
It is accessible within the same package.
d)
It is accessible throughout the program.
|
|
Sonal Yadav answered |
A local variable declared inside a method is only accessible within that method. It cannot be accessed from other methods or outside the method.
What will happen if a method with a void return type is called and the result is not assigned to any variable?- a)Compilation Error
- b)Runtime Error
- c)The program will execute normally.
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
What will happen if a method with a void return type is called and the result is not assigned to any variable?
a)
Compilation Error
b)
Runtime Error
c)
The program will execute normally.
d)
None of the above
|
|
Sonal Yadav answered |
When a method with a void return type is called, the method executes its statements but does not return a value. If the result is not assigned to any variable, it simply means that the result is not used, but the program will execute normally.
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int result = multiply(3, 4);
System.out.println(result);
} public static int multiply(int a, int b) {
return a * b;
}
}- a)12
- b)7
- c)Compilation Error
- d)Runtime Error
Correct answer is option 'A'. Can you explain this answer?
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int result = multiply(3, 4);
System.out.println(result);
}
public class MyClass {
public static void main(String[] args) {
int result = multiply(3, 4);
System.out.println(result);
}
public static int multiply(int a, int b) {
return a * b;
}
}
return a * b;
}
}
a)
12
b)
7
c)
Compilation Error
d)
Runtime Error
|
|
Sonal Yadav answered |
The 'multiply' method takes two integers as parameters and returns their product. The 'main' method calls the 'multiply' method with arguments 3 and 4, so the output will be the product of these numbers, which is 12.
What is method overloading in Java?- a)Defining multiple methods with the same name but different return types.
- b)Defining multiple methods with the same name but different parameter lists.
- c)Defining multiple methods with the same name but different access modifiers.
- d)Defining multiple methods with the same name but in different classes.
Correct answer is option 'B'. Can you explain this answer?
What is method overloading in Java?
a)
Defining multiple methods with the same name but different return types.
b)
Defining multiple methods with the same name but different parameter lists.
c)
Defining multiple methods with the same name but different access modifiers.
d)
Defining multiple methods with the same name but in different classes.
|
|
Sonal Yadav answered |
Method overloading in Java allows you to define multiple methods with the same name but different parameter lists. The compiler determines which method to call based on the arguments passed.
What is the output of the following code snippet?
public class Example {
public static void main(String[] args) {
String str = "Hello";
updateString(str);
System.out.println(str);
} public static void updateString(String s) {
s += " World";
}
}- a)Hello
- b)World
- c)Hello World
- d)Compilation error
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code snippet?
public class Example {
public static void main(String[] args) {
String str = "Hello";
updateString(str);
System.out.println(str);
}
public class Example {
public static void main(String[] args) {
String str = "Hello";
updateString(str);
System.out.println(str);
}
public static void updateString(String s) {
s += " World";
}
}
s += " World";
}
}
a)
Hello
b)
World
c)
Hello World
d)
Compilation error
|
|
Sonal Yadav answered |
The 'updateString' method in the code snippet takes a string parameter and concatenates " World" to it. However, the concatenation is performed on a copy of the reference to the original string. Hence, the modification is not reflected outside the method, and the output remains "Hello."
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int x = 10;
int y = modify(x);
System.out.println(x + y);
} public static int modify(int value) {
value += 5;
return value;
}
}- a)10
- b)15
- c)20
- d)Compilation Error
Correct answer is option 'C'. Can you explain this answer?
What will be the output of the following code?
public class MyClass {
public static void main(String[] args) {
int x = 10;
int y = modify(x);
System.out.println(x + y);
}
public class MyClass {
public static void main(String[] args) {
int x = 10;
int y = modify(x);
System.out.println(x + y);
}
public static int modify(int value) {
value += 5;
return value;
}
}
value += 5;
return value;
}
}
a)
10
b)
15
c)
20
d)
Compilation Error
|
|
Sonal Yadav answered |
The 'modify' method takes an integer as a parameter, adds 5 to it, and returns the modified value. In the 'main' method, the variable 'x' is assigned the value 10. When 'modify' is called with 'x' as an argument, 'x' itself is not modified, but the modified value (15) is returned and assigned to 'y'. The sum of 'x' (10) and 'y' (15) is 25, which is printed.
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