I am new, having a problem with my code.
I wonder how will I be able to print a variable after the if statements.
When I don't put the variable before the if statement it says that it's not initialized. here is the code:
import java.util.Scanner;
public class NameGenerator {
public static void main(String[] args) {
Scanner getInput = new Scanner(System.in);
System.out.println("What's the first letter of your first name? ");
String name = getInput.nextLine();
String xmas2;
if (name.equalsIgnoreCase ("a")) {
xmas2 = "Christmas";
} else if (name.equalsIgnoreCase ("b")) {
xmas2 = "Merry";
} else if (name.equalsIgnoreCase ("c")) {
xmas2 = "Santa";
} else if(name.equalsIgnoreCase ("d")) {
xmas2 = "Chocolate";
} else if(name.equalsIgnoreCase ("e")) {
xmas2 = "Tinsel";
} else if(name.equalsIgnoreCase ("f")) {
xmas2 = "Yule";
}
System.out.println("Hey " xmas2);
}
}
Thanks for the comment that pointed out to not put the string over and over
CodePudding user response:
First of all, you assign variable only 1 time " String xmas2 = null;"
import java.util.Scanner;
public class Practice {
public static void main(String[] args) {
Scanner getInput = new Scanner(System.in);
System.out.println("What's the first letter of your first name? ");
String name = getInput.nextLine();
String xmas2 = null;
if (name.equalsIgnoreCase("a")) {
xmas2 = "Christmas";
} else if (name.equalsIgnoreCase("b")) {
xmas2 = "Merry";
} else if (name.equalsIgnoreCase("c")) {
xmas2 = "Santa";
} else if (name.equalsIgnoreCase("d")) {
xmas2 = "Chocolate";
} else if (name.equalsIgnoreCase("e")) {
xmas2 = "Tinsel";
} else if (name.equalsIgnoreCase("f")) {
xmas2 = "Yule";
}
System.out.println("Hey " xmas2);
}
}
Output
What's the first letter of your first name?
a
Hey Christmas
CodePudding user response:
Note that class members are automatically initialized for you by the compiler but local variables are not. The compiler is issuing a warning to highlight the fact that you may encounter odd behavior if you do not populate the local variable prior to de-referencing it. Assign null to it initially and the warning goes away.
String xmas2 = null;
You can also move the logic here into a separate method which returns a value which you assign to initialize the variable. In the following I swap the if-else blocks for a switch and throw an Exception if the provided letter is not handled by the switch. Moving things into methods helps to define a unit of work.
public class Practice {
public static void main(String[] args) {
...
String xmas2 = interpretName(name);
...
}
private String interpretName(String name) {
String lower = name.toLowerCase();
switch (lower) {
case "a": return "Christmas";
case "b": return "Merry";
case "c": return "Santa";
case "d": return "Chocolate";
case "e": return "Tinsel";
case "f": return "Yule";
default: throw new IllegalArgumentException("Unhandled name: " name);
}
}
}
CodePudding user response:
In order to compile your code snippet, you need to initialize the variable xmas2 by using the assignment operator = after its declaration with one of the options below, before using the variable for any statements or calculations after that.
- Null:
String xmas2 = null; - Empty string:
String xmas2 = ""; - Or any String value:
String xmas2 = "Any string";
You may ask, why even the assignment String xmas2 = null; is able to fulfill the complaint Variable 'xmas2' might not have been initialized from the compiler. Does the String xmas2; declaration already do the same thing by automatically assigning it to a default value of null?
Then the answer is No. Java compiler will not automatically assign it - your local variable.
In Java, unassigned class member variables are automatically initialized to a default value, which in case a String, it should be assigned to null. Therefore if you have an unassigned class member, it will be able to compile successfully, due to the default value.
But unassigned local variables (Your xmas2 variable) are not assigned to any value until you manually assign it to a value (include null), therefore the compiler will give you a compile error.
