Java how to initialize String

Question

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

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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

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You can use below code to initialize size and set empty value to array of Strings  String   row   new String size   Arrays fill row

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String   arr     foo    bar      If you pass a string array to a method  do   myFunc arr     or do   myFunc new String     foo    bar

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You can always write it like this   String   errorSoon     Hello   World     For  int x 0 x lt 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 lt 2        System out println     errorSoon x       this will output those two     words  at the top hello and world at the bottom of hello

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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

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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

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You need to initialize errorSoon  as indicated by the error message  you have only declared it   String   errorSoon                        lt --declared statement String   errorSoon   new String 100       lt --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

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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

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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

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String   errorSoon      foo    bar       -- or --  String   errorSoon   new String 2   errorSoon 0     foo   errorSoon 1     bar

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String   args   new String    firstarg    secondarg    thirdarg

[java] Java: how to initialize String[]?

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