Java: how to initialize String[]?
JavaStringInitializationJava Problem Overview
Error
% javac StringTest.java
StringTest.java:4: variable errorSoon might not have been initialized
errorSoon[0] = "Error, why?";
Code
public class StringTest {
public static void main(String[] args) {
String[] errorSoon;
errorSoon[0] = "Error, why?";
}
}
Java Solutions
Solution 1 - Java
You need to initialize errorSoon
, as indicated by the error message, you have only declared it.
String[] errorSoon; // <--declared statement
String[] errorSoon = new String[100]; // <--initialized statement
You need to initialize the array so it can allocate the correct memory storage for the String
elements before you can start setting the index.
If you only declare the array (as you did) there is no memory allocated for the String
elements, but only a reference handle to errorSoon
, and will throw an error when you try to initialize a variable at any index.
As a side note, you could also initialize the String
array inside braces, { }
as so,
String[] errorSoon = {"Hello", "World"};
which is equivalent to
String[] errorSoon = new String[2];
errorSoon[0] = "Hello";
errorSoon[1] = "World";
Solution 2 - Java
String[] args = new String[]{"firstarg", "secondarg", "thirdarg"};
Solution 3 - Java
String[] errorSoon = { "foo", "bar" };
-- or --
String[] errorSoon = new String[2];
errorSoon[0] = "foo";
errorSoon[1] = "bar";
Solution 4 - Java
In Java 8 we can also make use of streams e.g.
String[] strings = Stream.of("First", "Second", "Third").toArray(String[]::new);
In case we already have a list of strings (stringList
) then we can collect into string array as:
String[] strings = stringList.stream().toArray(String[]::new);
Solution 5 - Java
I believe you just migrated from C++, Well in java you have to initialize a data type(other then primitive types and String is not a considered as a primitive type in java ) to use them as according to their specifications if you don't then its just like an empty reference variable (much like a pointer in the context of C++).
public class StringTest {
public static void main(String[] args) {
String[] errorSoon = new String[100];
errorSoon[0] = "Error, why?";
//another approach would be direct initialization
String[] errorsoon = {"Error , why?"};
}
}
Solution 6 - Java
String[] arr = {"foo", "bar"};
If you pass a string array to a method, do:
myFunc(arr);
or do:
myFunc(new String[] {"foo", "bar"});
Solution 7 - Java
String[] errorSoon = new String[n];
With n being how many strings it needs to hold.
You can do that in the declaration, or do it without the String[] later on, so long as it's before you try use them.
Solution 8 - Java
You can always write it like this
String[] errorSoon = {"Hello","World"};
For (int x=0;x<errorSoon.length;x++) // in this way u create a for loop that would like display the elements which are inside the array errorSoon.oh errorSoon.length is the same as errorSoon<2
{
System.out.println(" "+errorSoon[x]); // this will output those two words, at the top hello and world at the bottom of hello.
}
Solution 9 - Java
You can use below code to initialize size and set empty value to array of Strings
String[] row = new String[size];
Arrays.fill(row, "");
Solution 10 - Java
String Declaration:
String str;
String Initialization
String[] str=new String[3];//if we give string[2] will get Exception insted
str[0]="Tej";
str[1]="Good";
str[2]="Girl";
String str="SSN";
We can get individual character in String:
char chr=str.charAt(0);`//output will be S`
If I want to to get individual character Ascii value like this:
System.out.println((int)chr); //output:83
Now i want to convert Ascii value into Charecter/Symbol.
int n=(int)chr;
System.out.println((char)n);//output:S
Solution 11 - Java
String[] string=new String[60];
System.out.println(string.length);
> it is initialization and getting the STRING LENGTH code in very simple way for beginners