How to convert String to long in Java

Question

I got a simple question in Java  How can I convert a String that was obtained by Long toString   to long

User · Answer

It s quite simple  use Long valueOf String s    For example   String s  long l   Scanner sc new Scanner System in   s sc next    l Long valueOf s   System out print l     You re done

User · Answer

Use Long parseLong       Long parseLong  0   10            returns 0L  Long parseLong  473   10          returns 473L  Long parseLong  -0   10           returns 0L  Long parseLong  -FF   16          returns -255L  Long parseLong  1100110   2       returns 102L  Long parseLong  99   8            throws a NumberFormatException  Long parseLong  Hazelnut   10     throws a NumberFormatException  Long parseLong  Hazelnut   36     returns 1356099454469L  Long parseLong  999               returns 999L

User · Answer

public class StringToLong       public static void main  String   args              String s    fred         do this if you want an exception        String s    100          try            long l   Long parseLong s            System out println  long l       l           catch  NumberFormatException nfe             System out println  NumberFormatException      nfe getMessage

User · Answer

Long valueOf String s  - obviously due care must be taken to protect against non-numbers if that is possible in your code

User · Answer

For those who switched to Kotlin just use string toLong   That will call Long parseLong string  under the hood

User · Answer

There are a few ways to convert String to long  1  long l   Long parseLong  quot 200 quot         String numberAsString    quot 1234 quot   long number   Long valueOf numberAsString  longValue        String numberAsString    quot 1234 quot   Long longObject   new Long numberAsString   long number   longObject longValue     We can shorten to  String numberAsString    quot 1234 quot   long number   new Long numberAsString  longValue     Or just long number   new Long  quot 1234 quot   longValue      Using Decimal format   String numberAsString    quot 1234 quot   DecimalFormat decimalFormat   new DecimalFormat  quot   quot    try       long number   decimalFormat parse numberAsString  longValue        System out println  quot The number is   quot    number     catch  ParseException e        System out println numberAsString    quot  is not a valid number  quot

User · Answer

In case you are using the Map with out generic  then you need to convert the value into String and then try to convert to Long  Below is sample code      Map map   new HashMap         map put  name    John        map put  time    9648512236521        map put  age    25         long time   Long valueOf  String map get  time    longValue         int age   Integer valueOf  String   map get  aget    intValue        System out println time       System out println age

User · Answer

To convert a String to a Long  object   use Long valueOf String s  longValue     See link

User · Answer

The best approach is Long valueOf str  as it relies on Long valueOf long  which uses an internal cache making it more efficient since it will reuse if needed the cached instances of Long going from -128 to 127 included      Returns a Long instance representing the specified long value  If a   new Long instance is not required  this method should generally be   used in preference to the constructor Long long   as this method is   likely to yield significantly better space and time performance by   caching frequently requested values  Note that unlike the   corresponding method in the Integer class  this method is not required   to cache values within a particular range    Thanks to auto-unboxing allowing to convert a wrapper class s instance into its corresponding primitive type  the code would then be   long val   Long valueOf str     Please note that the previous code can still throw a NumberFormatException if the provided String doesn t match with a signed long   Generally speaking  it is a good practice to use the static factory method valueOf str  of a wrapper class like Integer  Boolean  Long      since most of them reuse instances whenever it is possible making them potentially more efficient in term of memory footprint than the corresponding parse methods or constructors     Excerpt from Effective Java Item 1 written by Joshua Bloch      You can often avoid creating unnecessary objects by using static   factory methods  Item 1  in preference to constructors on immutable   classes that provide both  For example  the static factory method   Boolean valueOf String  is almost always preferable to the   constructor Boolean String   The constructor creates a new object   each time it   s called  while the static factory method is never   required to do so and won   t in practice

[java] How to convert String to long in Java?

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