Java silver se17 Quizzes 3

1. Argument passing
With the command `java A He llo "He llo" 1 2 3`, the argument `i` of the main method stores the following 6 strings:
i={"He","llo","He llo","1","2","3"}

2. replace method processing
`i = replace(4, i);` is executed.
Creation of new array: The value 4 from `i` is used as the array size, creating a new array `l` of size 4.
Element manipulation: The loop processes the first 4 elements of the original array `i`. `j[k].replace(" ", "")` removes all half-width spaces from each element.
`i` in the main method is replaced by this array: {"He", "llo", "Hello", "1"}.

3. Output processing
The for loop in the main method processes the new 4-element array in order. After printing each element, `System.out.print(" ");` always outputs a half-width space.

Class E inherits from C, and ultimately bears the obligation to implement the abstract method of Interface A.
Abstract method defined in `interface A`: `A a(A a);`
`abstract class C` does not implement this method, but since it is an abstract class, that is fine.
`class E` is a concrete class, so it must implement this abstract method.

2. Implementation in Class E
Required signature (inherited from A): `A a(A a)`
Defined signature (defined in E): `A a(E e)`
The method in E changes the argument type from A to its subclass E. This is considered an overload (definition of a new method), not an override.

3. Occurrence of compilation error
Class E does not provide a matching implementation for the inherited abstract method `A a(A a)`. Nevertheless, since E is a concrete class not declared as abstract, the following compilation error occurs:
Error: Class E must implement the abstract method `A a(A a)` inherited from the superclass, or declare itself as abstract.

Checking other classes
A, B: Only declarations, no problem.

D: `default A a(D d)` is an overload because the signature differs from `A.a(A a)`, but since it is an interface, there is no problem.

C: Since it is an abstract class, it does not have to implement `A.a(A a)`. Also, `C b(B b)` is a valid override of `B.b(B b)` (the return type can be changed to B's subclass C due to covariant return types).

1. `if((a&b++)>(a|++b)&b-->0)`
Evaluation of left side: `(a & b++) > (a | ++b)`
`a & b++` : `1001 & 0101` → `1`. `b` becomes `0110` due to post-increment.
`a | ++b` : `b` becomes `0111` due to pre-increment.
`1001 | 0111` → `1111`.
Comparison: `1 > 1111` → `false`.

Evaluation of logical AND (&): `&` does not perform short-circuit evaluation, so the right side is also evaluated.
Right side `b-->0`:
`0111 > 0` → `true`.
`b` becomes `0110` due to post-decrement.

Final result `false & true` → `false` → The first if statement is not executed.

2. `if((--a^b)<(~b)&&b-->0)`
Evaluation of left side: `(--a^b) < (~b)`
`--a^b`: `a` becomes `1000` due to pre-decrement. `1000 ^ 0110` → `1110`.
`~b`: Bitwise complement of `b=110` → `1...1001` (Negative number because the 32nd bit is 1).
Comparison: `1110` (positive) < (negative) → `false`.

Evaluation of logical AND (&&):
The result of the left side is `false`. Since `&&` performs short-circuit evaluation, the right side `b-->0` is not evaluated.

→ The 2nd if statement is not executed, and the value of `b` remains `110`.

3. Evaluation of 3rd if statement and output
`if((-a>>2)<(-a>>>2)|a<(b<<1)|a++>3||a-->2)`
Evaluation of logical OR (|) (no short-circuit): `|` does not short-circuit, so the first three conditions are evaluated in order.

`(-a >> 2) < (-a >>> 2)`
Signed right shift of `-a` (`1...1000`): `-a >> 2` → `1...10` (negative number).
Zero-fill right shift: `-a >>> 2` → `001...10` (positive number).
Comparison: (negative) < (positive) → `true`.

Since the first condition is `true`, the overall result will be `true`, but the remaining conditions connected by `|` are still evaluated.

`a < (b << 1)`
Left shift of `b` (`110`): `b << 1` → `1100`.
Comparison: `1000 < 1100` → `true`.

`a++ > 3`
`1000 > 3` → `true`.
`a` becomes `1001` due to post-increment.

Evaluation of logical OR (||) (with short-circuit):
At `(true | true | true)`, the result is already `true`. Since `||` performs short-circuit evaluation, the right side `a-->2` is not evaluated.
→ The 3rd if statement is executed.

5. Output `System.out.print(a + b);`
Final value of `a`: `1001` = 2^3 + 1 = 9. Final value of `b`: `110` = 2^4 + 2^1 = 6.
`a + b` = 9 + 6 = 15.

The reason this code causes a compilation error lies in the declaration of the `toString()` method within interface B.
Under Java rules, an interface cannot redefine public methods of the Object class (such as `toString()`, `equals()`, `hashCode()`, etc.) as default methods.
Due to this rule, the declaration `default String toString()` in interface B violates the Java language specification, so the compiler generates an error when attempting to compile this code.

1. Determination of input values
Through the double loop in the main method, the following three integer values are passed sequentially to `new B().access()`:
1: (int)1.9, 0: (int)0.5, 5: (int)5.01

Inside the `access(int i)` method, the key for the switch statement is `i * 2`.
1st time: 1 * 2 = 2, 2nd time: 0 * 2 = 0, 3rd time: 5 * 2 = 10

2. Behavior of switch statement (Determining output)
Each case value in the switch statement is determined at compile time.
`case A.i`: value is 2, `case 3*3`: value is 9, `case 1/3`: integer division, so value is 0, `case 3`: value is 3.

1st time: When i * 2 = 2 (Output: 12)
Matches `case 2` (case A.i). Processing starts.
Since there is no processing immediately after `case 2`, it falls through to the next `case 9`.
`case 9` processing `System.out.print(1);` is executed, printing 1.
Next, it falls through to `default`.
`default` processing `System.out.print(2);` is executed, printing 2.
Reaches `break a;`, exits the labeled switch statement entirely, and ends the method.

2nd time: When i * 2 = 0 (Output: 34)
Matches `case 0`, prints 3.
Next, falls through to `case 3` and prints 4.

3rd time: When i * 2 = 10 (Output: 2)
Matches no case label, so processing jumps to `default`.
`default` processing `System.out.print(2);` is executed, printing 2.
Reaches `break a;`, ending the switch statement.

Compilation Error 1: Array size specification
`short o=2;long p=2;`
`byte[]k[]=new byte[o*p][o*(int)p];` // ← Error here
`o` is short type, `p` is long type.
In the calculation `o * p`, the short type `o` is promoted (type promotion) to long, so the result is long type `4L`.
Array sizes cannot be specified with long type; they must be int type.
Therefore, it is interpreted like `new byte[4L][4]`, causing a compilation error in the size specification of the first dimension.

Compilation Error 2: Initialization block of for loop
`for(byte i=0,int j=0;i<k.length&j<k.length;i++,j++,dot()) {` // ← Error here
When declaring multiple variables in the initialization block of a for loop, they must all be of the same type.
This code attempts to declare two different types, byte type `i` and int type `j`, simultaneously, resulting in a syntax error and failure to compile. Correctly, types must match, like `int i=0, j=0;` or `byte i=0, b=0;`.

1. Initialization and Capacity
`j` is initialized with `new StringBuilder(17)`, so its capacity (`j.capacity()`) is 17.
`i` is initialized with `new StringBuilder()`, so its default capacity (`i.capacity()`) is typically 16.

2. Content manipulation of `i`
`i.append("ab".repeat(8)).append("cd".repeat(1));`
`"ab".repeat(8)` is 16 characters: "abababababababab".

2 characters "cd" are appended, making `i`'s content "ababababababababcd" (length 18).

Since length 18 exceeds the initial capacity of 16, `i`'s capacity is expanded (to be 18 or more).

Current: `i` content: "ababababababababcd", Capacity: 34

3. Replacement and Reversal
`i.replace(1, 16, "").reverse();`
`i.replace(1, 16, "")`:
15 characters from index 1 to 16 ('b' to 'b') are deleted.

Original string: a bababababababab cd
From this, 'a' at index 0, 'c' at index 16, and 'd' at index 17 remain.
`i`'s content becomes "acd".

`i.reverse()`: `i`'s content becomes "dca".

4. Final character manipulation
`i.deleteCharAt(1).insert(1,'e');`
`i.deleteCharAt(1)`: Character 'c' at index 1 is deleted. `i`'s content becomes "da".

`i.insert(1, 'e')`: 'e' is inserted at index 1. `i`'s content becomes "dea".

5. Condition check and output
`if(i.capacity()<j.capacity())System.out.print(i.reverse());`
`else System.out.println(i);`
`i.capacity()` is 34. `j.capacity()` is 17.
34 < 17 is false.
→ The `else` block is executed, and the current content of `i`, "dea", is output with a newline.

A: Incorrect (Compilation error)
Control variable: `c` is Character (char) type.
Error reason: Since char type can take many values other than 'a' or 'b', the lack of a default block fails to satisfy exhaustiveness, causing a compilation error.

B: Incorrect (Compilation error)
Error reason: Control variable `i` is long type. `long` is not a valid type for a switch control variable, causing a compilation error. Usable types are int, byte, short, char, their wrapper classes (Integer, etc.), String, and enum.

C: Correct (Compilable)
Control variable: `b` is Byte type, which is valid for a switch control variable.
Exhaustiveness: Since a default block is written, exhaustiveness is satisfied.
Syntax: It is correct syntax for a switch expression, and compilation succeeds. `yield` is used to return values inside `{}`.

D: Incorrect (Compilation error)
Error reason 1: Control variable `f` is float type. `float` is not a valid type for a switch control variable, causing a compilation error.

A: `abstract` (abstract) class is a modifier that assumes inheritance.
`final` class is a modifier that prohibits inheritance. These two contradictory modifiers cannot be specified on a class simultaneously.

B: Since interfaces are intended to be implemented, the `final` modifier cannot be attached.

C: Looks correct at first glance, but the call to `A.super.a()` is problematic.
When calling a default method with the syntax `InterfaceName.super.methodName()`, that `InterfaceName` must be an interface that the calling class directly implements (`implements`).
In this code, class E directly implements B, but A is an interface that B inherits from, not one that E implements directly (it is an indirect superinterface). Therefore, the call `A.super.a()` results in a compilation error.

D: You cannot attach `protected` or `final` modifiers to interface methods.

E: `public A(){}` is a constructor description.
Interfaces cannot be instantiated, so they cannot have constructors.

F: Fields (variables) declared in an interface are automatically treated as `public static final` (constants).
`final` fields must be initialized at declaration, but `int i;` is uninitialized, causing a compilation error. Correctly, you must assign a value like `int i = 10;`.