1Z0-829 Controlling Program Flow Practice Test 3 — 15 Questions

Question 1

Which THREE statements are true about the enhanced for loop in Java?

Accepted Answer

The enhanced for loop internally uses an iterator for collections.. The enhanced for loop (for-each) over a collection internally uses an Iterator obtained from the collection's iterator() method. This is specified in the Java Language Specification (JLS 14.14.2) and is the mechanism that allows sequential access to each element without exposing the iterator directly.

Answer

If the underlying collection is modified during iteration, the loop will continue without error.

Answer

The loop variable must be declared final in the loop header.

Question 2

Which TWO of the following are valid forms of the switch statement/expression in Java?

Accepted Answer

switch(x) { case 1: System.out.println("one"); break; }. It uses the traditional switch statement syntax with a colon after the case label, a statement to execute, and a break statement to prevent fall-through. This is a valid and long-standing form of the switch statement in Java.

Answer

switch(x) { case 1 -> "one"; }

Answer

String r = switch(x) { case 1: break "one"; };

Answer

switch(x) { case 1: yield "one"; }

Question 3

What is the output when this code is executed?

```java
int x = 2;
switch (x) {
    case 1:
        System.out.print("One ");
    case 2:
        System.out.print("Two ");
    case 3:
        System.out.print("Three ");
        break;
    default:
        System.out.print("Default");
}
```

Accepted Answer

Two Three. The switch statement matches the value 2, so it executes the case 'Two' block, which prints 'Two ' and then falls through to the next case because there is no break statement. The case 'Three' block prints 'Three ' and then breaks, exiting the switch. The default case is skipped because a matching case was found and the break in case 'Three' prevents further fall-through. Thus, the output is 'Two Three '.

Answer

Two

Answer

Two Three Default

Answer

One Two Three

Question 4

Which change fixes the exception?

Accepted Answer

Change line 4 to: for (int i = 0; i < arr.length; i++). It changes the loop to start at index 0 and continue while i < arr.length, which correctly iterates over all valid indices (0 to arr.length-1). The original code likely used i <= arr.length, causing an ArrayIndexOutOfBoundsException when accessing arr[i] at i = arr.length. By using i < arr.length, the loop stops before reaching the out-of-bounds index.

Answer

Change line 4 to: for (int i = 1; i <= arr.length; i++)

Answer

Change line 4 to: for (int i = 0; i <= arr.length-1; i++)

Answer

Change line 5 to: System.out.println(arr[i-1]);

Question 5

What is the likely cause of this compilation error?
```java
String day = "Monday";
int dayNum = switch (day) {
    case 1 -> 1;
    case 2 -> 2;
    default -> 0;
};
System.out.println(dayNum);
```

Accepted Answer

The switch selector is a String but case labels are int constants.. In a Java switch expression, the selector type and case label types must be compatible. If the selector is a String, the case labels must be String constants; int constants would cause a compilation error due to type mismatch. Options A, B, and C describe syntax issues that are not the primary cause here.

Answer

The switch expression uses arrow syntax but the keys are not unique.

Answer

The switch expression is missing a default case.

Answer

The switch expression is not assigned to a variable.

Question 6

Given: for(int i=0; i<3; i++) { for(int j=0; j<3; j++) { if(i==1 && j==1) break; } } What is the value of i after the outer loop completes?

Accepted Answer

3. The outer loop variable `i` increments from 0 to 2. When `i` is 1 and `j` is 1, the `break` statement exits only the inner `for` loop, not the outer loop. The outer loop continues to its next iteration, incrementing `i` to 2, then to 3. When `i` becomes 3, the outer loop condition `i < 3` fails, so the loop terminates. Thus, after the outer loop completes, `i` is 3.

Answer

1

Answer

2

Answer

4

Question 7

Given: Object obj = "Hello"; String result = switch(obj) { case String s -> "String of length " + s.length(); case Integer i -> "Integer"; default -> "Unknown"; }; What is result?

Accepted Answer

"String of length 5". The switch expression uses pattern matching with type patterns. The runtime type of obj is String, so the first case matches, binding the value to s and executing the expression "String of length " + s.length(). Since s is "Hello", s.length() returns 5, so result is "String of length 5". Option C is correct.

Answer

"Unknown"

Answer

Compilation error

Answer

"Integer"

Question 8

Given: outer: for(int i=0; i<3; i++) { for(int j=0; j<3; j++) { if(j==1) continue outer; } } How many times does the innermost loop body execute?

Accepted Answer

3. The outer loop runs with i=0,1,2. For each i, the inner loop starts j=0, executes the body once, then j=1 triggers 'continue outer', which skips the rest of the inner loop and increments i. Thus the inner loop body executes exactly once per outer iteration, for a total of 3 times.

Answer

0

Answer

9

Answer

6

Question 9

Given: int x=0; do { x++; } while(x<5); How many times does the loop body execute?

Accepted Answer

5. The do-while loop executes the body first, then checks the condition. The variable x starts at 0, so initially the body executes with x=0, then x increments to 1. The loop continues as long as x < 5. The body executes for x = 0, 1, 2, 3, 4 — a total of 5 times. When x becomes 5, the condition x < 5 is false, so the loop stops.

Answer

4

Answer

6

Answer

0

Question 10

Given: int a=5, b=10; String result = (a > b) ? "greater" : (a < b) ? "less" : "equal"; What is result?

Accepted Answer

"less". The ternary operator is right-associative, so the expression is parsed as `a > b ? "greater" : (a < b ? "less" : "equal")`. Since `a=5` and `b=10`, `a > b` is false, so the second operand of the outer ternary is evaluated: `a < b` is true, so the inner ternary returns `"less"`. The code compiles and runs without error, producing `"less"`.

Answer

"equal"

Answer

"greater"

Answer

Compilation error

Question 11

Given: Object obj = null; String s = switch(obj) { case null -> "null"; case String str -> str; default -> "other"; }; What is the result?

Accepted Answer

"null". In Java 14+, switch expressions can handle null values directly using a 'case null' label. Since obj is null, the switch matches the 'case null' branch and returns the string "null". Option D is correct because the code compiles and executes without error, producing "null".

Answer

Compilation error

Answer

NullPointerException

Answer

"other"

Question 12

Given: int i=0; outer: while(i<3) { for(int j=0; j<3; j++) { if(j==1) break outer; } i++; } What is the value of i after the outer loop?

Accepted Answer

0. (0) because the `break outer` statement immediately terminates the outer `while` loop when `j` equals 1 during the first iteration of the inner `for` loop. Since `i` is incremented only after the inner loop completes, and the inner loop never finishes its first iteration, `i` remains 0.

Answer

1

Answer

2

Answer

3

Question 13

Given: for(int i=0; i<10; i++) { if(i%2==0) continue; System.out.print(i+" "); } What is the output?

Accepted Answer

1 3 5 7 9. The loop iterates from 0 to 9, and the `continue` statement skips the rest of the loop body when the condition `i % 2 == 0` is true (i.e., for even numbers). Only odd numbers (1, 3, 5, 7, 9) are printed, each followed by a space.

Answer

No output

Answer

0 2 4 6 8

Answer

0 1 2 3 4 5 6 7 8 9

Question 14

Given: int x=1; switch(x) { case 1: System.out.print("A"); case 2: System.out.print("B"); break; default: System.out.print("C"); } What is printed?

Accepted Answer

AB. The switch statement does not have a break after case 1, so execution falls through to case 2 after printing 'A'. Case 2 prints 'B' and then hits the break, exiting the switch. The default case is skipped because a matching case was found. Thus, 'AB' is printed.

Answer

C

Answer

ABC

Answer

A

Question 15

Which two statements about the break statement are true?

Accepted Answer

break without a label terminates the innermost loop or switch.. An unlabeled break statement terminates the innermost enclosing loop (for, while, do-while) or switch statement, as defined by the Java Language Specification. Option E is correct because break is valid inside a switch statement to exit the switch block. Option A is incorrect; break cannot be used inside an if statement unless that if is inside a loop or switch, but the break itself must be directly inside the loop/switch. Option C is incorrect because break can also be used in a switch, not only in loops. Option D is incorrect; a labeled break terminates the statement with the matching label, which could be any labeled statement (block, loop, or switch), not necessarily the outermost loop or switch.

Answer

break can be used inside an if statement.

Answer

break can only be used inside a loop.

Answer

break with a label terminates the outermost matching loop or switch.