Checkpoint 3.59 Assume the variables name1 and name2 reference two different String objects, containing different strings. Write code that displays the strings referenced by these variables in alphabetical order. System.out.println(name1); System.out.println(name2); } else { System.out.println(name2); System.out.println(name1); } Fill in the blank. + namel + alphabetically.");| + comes before

Microsoft Visual C#
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ISBN:9781337102100
Author:Joyce, Farrell.
Publisher:Joyce, Farrell.
Chapter8: Advanced Method Concepts
Section: Chapter Questions
Problem 8RQ
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java

### Checkpoint 3.59

#### Problem Statement:
Assume the variables `name1` and `name2` reference two different `String` objects, containing different strings. Write code that displays the strings referenced by these variables in alphabetical order.

```java
________ 
{ 
    System.out.println(name1); 
    System.out.println(name2); 
} 
else 
{ 
    System.out.println(name2); 
    System.out.println(name1); 
}
```

#### Task:
Fill in the blank.

#### Incorrect Attempt:
An attempt was made to fill in the blank with: 
```java
+ " comes before " + name1 + " alphabetically.");
``` 
This resulted in an incorrect response, as marked by the red 'X' and the message "Incorrect. Try again."

### Solution Explanation:

To correctly display `name1` and `name2` in alphabetical order, you need to compare the two strings using the `compareTo` method of the `String` class. 

Here is the correct code:

```java
if (name1.compareTo(name2) < 0)
{
    System.out.println(name1);
    System.out.println(name2);
}
else
{
    System.out.println(name2);
    System.out.println(name1);
}
```

### Detailed Explanation:
1. **String Comparison:** 
   - `name1.compareTo(name2) < 0` checks if `name1` is alphabetically before `name2`. The `compareTo` method returns:
     - A negative integer if `name1` comes before `name2`.
     - Zero if `name1` and `name2` are equal.
     - A positive integer if `name1` comes after `name2`.

2. **If-Else Structure:**
   - If the condition `name1.compareTo(name2) < 0` is true, `name1` is printed first followed by `name2`.
   - If the condition is false, `name2` is printed first followed by `name1`.

This ensures that the strings are always printed in alphabetical order.

Remember to always test the logic with different strings to ensure correctness.
Transcribed Image Text:### Checkpoint 3.59 #### Problem Statement: Assume the variables `name1` and `name2` reference two different `String` objects, containing different strings. Write code that displays the strings referenced by these variables in alphabetical order. ```java ________ { System.out.println(name1); System.out.println(name2); } else { System.out.println(name2); System.out.println(name1); } ``` #### Task: Fill in the blank. #### Incorrect Attempt: An attempt was made to fill in the blank with: ```java + " comes before " + name1 + " alphabetically."); ``` This resulted in an incorrect response, as marked by the red 'X' and the message "Incorrect. Try again." ### Solution Explanation: To correctly display `name1` and `name2` in alphabetical order, you need to compare the two strings using the `compareTo` method of the `String` class. Here is the correct code: ```java if (name1.compareTo(name2) < 0) { System.out.println(name1); System.out.println(name2); } else { System.out.println(name2); System.out.println(name1); } ``` ### Detailed Explanation: 1. **String Comparison:** - `name1.compareTo(name2) < 0` checks if `name1` is alphabetically before `name2`. The `compareTo` method returns: - A negative integer if `name1` comes before `name2`. - Zero if `name1` and `name2` are equal. - A positive integer if `name1` comes after `name2`. 2. **If-Else Structure:** - If the condition `name1.compareTo(name2) < 0` is true, `name1` is printed first followed by `name2`. - If the condition is false, `name2` is printed first followed by `name1`. This ensures that the strings are always printed in alphabetical order. Remember to always test the logic with different strings to ensure correctness.
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