How do I convert a String to an int in Java

Question

How can I convert a String to an int in Java   My String contains only numbers  and I want to return the number it represents   For example  given the string  1234  the result should be the number 1234

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You can have your own implementations for this  like  public class NumericStringToInt        public static void main String   args            String str    quot 123459 quot            int num   stringToNumber str           System out println  quot Number of  quot    str    quot  is   quot    num              private static int stringToNumber String str             int num   0          int i   0          while  i  lt  str length                  char ch   str charAt i               if  ch  lt  48    ch  gt  57                  throw new NumberFormatException  quot  quot    ch               num   num   10   Character getNumericValue ch               i                      return num

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Use this method   public int ConvertStringToInt String number        int num   0       try           int newNumber   Integer ParseInt number           num   newNumber        catch Exception ex            num   0          Log i  Console   ex toString             return num

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Use Integer parseInt yourString    Remember the following things   Integer parseInt  1            ok  Integer parseInt  -1           ok  Integer parseInt   1           ok  Integer parseInt   1           Exception  blank space   Integer parseInt  2147483648       Exception  Integer is limited to a maximum value of 2 147 483 647   Integer parseInt  1 1       Exception    or   or whatever is not allowed   Integer parseInt         Exception  not 0 or something   There is only one type of exception  NumberFormatException

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For example  here are two ways   Integer x   Integer valueOf str      or int y   Integer parseInt str     There is a slight difference between these methods     valueOf returns a new or cached instance of java lang Integer parseInt returns primitive int     The same is for all cases  Short valueOf parseShort  Long valueOf parseLong  etc

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Kindly use NumberUtils to parse integers from a string   This function could also handle the exception when the given string is too long  We could also give default values   Here is the example code  NumberUtils toInt  quot 00450 quot    NumberUtils toInt  quot 45464646545645400000 quot    NumberUtils toInt  quot 45464646545645400000 quot   0      Where 0 is the default value   output  450 0 0

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Some of the ways to convert String into Int are as follows    You can use Integer parseInt     String test    4568   int new   Integer parseInt test    Also you can use Integer valueOf     String test    4568   int new   Integer valueOf test

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Integer decode  You can also use public static Integer decode String nm  throws NumberFormatException   It also works for base 8 and 16      base 10 Integer parseInt  12           12 - int Integer valueOf  12            12 - Integer Integer decode  12             12 - Integer    base 8    10  0 1     7 10 11 12  Integer parseInt  12   8       10 - int Integer valueOf  12   8        10 - Integer Integer decode  012            10 - Integer    base 16    18  0 1     F 10 11 12  Integer parseInt  12  16       18 - int Integer valueOf  12  16        18 - Integer Integer decode   12            18 - Integer Integer decode  0x12           18 - Integer Integer decode  0X12           18 - Integer    base 2 Integer parseInt  11  2        3 - int Integer valueOf  11  2         3 - Integer   If you want to get int instead of Integer you can use    Unboxing   int val   Integer decode  12      intValue     Integer decode  12   intValue

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By using this method you can avoid errors   String myString    1234   int myInt  if Integer parseInt myString   out myInt

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public static int parseInt String s throws NumberFormatException  You can use Integer parseInt   to convert a String to an int   Convert a String   20   to a primitive int   String n    20   int r   Integer parseInt n      Returns a primitive int System out println r        Output-20   If the string does not contain a parsable integer  it will throw NumberFormatException   String n    20I      Throws NumberFormatException int r   Integer parseInt n   System out println r     public static Integer valueOf String s throws NumberFormatException  You can use Integer valueOf    In this it will return an Integer object   String n    20   Integer r   Integer valueOf n      Returns a new Integer   object  System out println r        Output-20   References https   docs oracle com en

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The two main ways to do this are using the method valueOf   and method parseInt   of the Integer class  Suppose you are given a String like this String numberInString    quot 999 quot    Then you can convert it into integer by using int numberInInteger   Integer parseInt numberInString    And alternatively  you can use int numberInInteger   Integer valueOf numberInString    But the thing here is  the method Integer valueOf   has the following implementation in Integer class  public static Integer valueOf String var0  int var1  throws NumberFormatException       return parseInt var0  var1      As you can see  the Integer valueOf   internally calls Integer parseInt   itself  Also  parseInt   returns an int  and valueOf   returns an Integer

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As I write on GitHub   public class StringToInteger       public static void main String   args            assert parseInt  123      Integer parseInt  123            assert parseInt  -123      Integer parseInt  -123            assert parseInt  0123      Integer parseInt  0123            assert parseInt   123      Integer parseInt   123                          Parse a string to integer                param s the string         return the integer value represented by the argument in decimal          throws NumberFormatException if the   code string  does not contain a parsable integer              public static int parseInt String s            if  s    null                throw new NumberFormatException  null                      boolean isNegative   s charAt 0      -           boolean isPositive   s charAt 0                  int number   0          for  int i   isNegative   1   isPositive   1   0  length   s length    i  lt  length    i                if   Character isDigit s charAt i                      throw new NumberFormatException  s     s                             number   number   10   s charAt i  -  0                     return isNegative   -number   number

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In programming competitions  where you re assured that number will always be a valid integer  then you can write your own method to parse input  This will skip all validation related code  since you don t need any of that  and will be a bit more efficient    For valid positive integer   private static int parseInt String str        int i  n   0       for  i   0  i  lt  str length    i              n    10          n    str charAt i  - 48            return n     For both positive and negative integers   private static int parseInt String str        int i 0  n 0  sign 1      if  str charAt 0      -             i   1          sign   -1            for   i lt str length    i              n    10          n    str charAt i  - 48            return sign n     If you are expecting a whitespace before or after these numbers  then make sure to do a str   str trim   before processing further

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Currently I m doing an assignment for college  where I can t use certain expressions  such as the ones above  and by looking at the ASCII table  I managed to do it  It s a far more complex code  but it could help others that are restricted like I was  The first thing to do is to receive the input  in this case  a string of digits  I ll call it String number  and in this case  I ll exemplify it using the number 12  therefore String number    quot 12 quot   Another limitation was the fact that I couldn t use repetitive cycles  therefore  a for cycle  which would have been perfect  can t be used either  This limits us a bit  but then again  that s the goal  Since I only needed two digits  taking the last two digits   a simple charAtsolved it      Obtaining the integer values of the char 1 and 2 in ASCII  int semilastdigitASCII   number charAt number length   - 2    int lastdigitASCII   number charAt number length   - 1    Having the codes  we just need to look up at the table  and make the necessary adjustments   double semilastdigit   semilastdigitASCII - 48      A quick look  and -48 is the key  double lastdigit   lastdigitASCII - 48   Now  why double  Well  because of a really  quot weird quot  step  Currently we have two doubles  1 and 2  but we need to turn it into 12  there isn t any mathematic operation that we can do  We re dividing the latter  lastdigit  by 10 in the fashion 2 10   0 2  hence why double  like this   lastdigit   lastdigit   10   This is merely playing with numbers  We were turning the last digit into a decimal  But now  look at what happens   double jointdigits   semilastdigit   lastdigit     1 0   0 2   1 2  Without getting too into the math  we re simply isolating units the digits of a number  You see  since we only consider 0-9  dividing by a multiple of 10 is like creating a  quot box quot  where you store it  think back at when your first grade teacher explained you what a unit and a hundred were   So   int finalnumber    int   jointdigits 10      Be sure to use parentheses  quot    quot   And there you go  You turned a String of digits  in this case  two digits   into an integer composed of those two digits  considering the following limitations   No repetitive cycles No  quot Magic quot  Expressions such as parseInt

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This is a complete program with all conditions positive and negative without using a library  import java util Scanner    public class StringToInt        public static void main String args              String inputString          Scanner s   new Scanner System in           inputString   s nextLine             if   inputString matches     -    0-9        0-9                     System out println  Not a Number                      else               Double result2   getNumber inputString               System out println  result       result2                         public static Double getNumber String number            Double result   0 0          Double beforeDecimal   0 0          Double afterDecimal   0 0          Double afterDecimalCount   0 0          int signBit   1          boolean flag   false           int count   number length            if  number charAt 0      -                 signBit   -1              flag   true                    else if  number charAt 0                        flag   true                    for  int i   0  i  lt  count  i                  if  flag  amp  amp  i    0                    continue                            if  afterDecimalCount    0 0                    if  number charAt i  -        0                        afterDecimalCount                                      else                       beforeDecimal   beforeDecimal   10    number charAt i  -  0                                                else                   afterDecimal   afterDecimal   10   number charAt i  -   0                    afterDecimalCount   afterDecimalCount   10                                  if  afterDecimalCount    0 0                afterDecimal   afterDecimal   afterDecimalCount              result   beforeDecimal   afterDecimal                    else               result   beforeDecimal                    return result   signBit

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You can use this code also  with some precautions    Option  1  Handle the exception explicitly  for example  showing a message dialog and then stop the execution of the current workflow  For example   try               String stringValue    1234               From String to Integer         int integerValue   Integer valueOf stringValue               Or         int integerValue   Integer ParseInt stringValue               Now from integer to back into string         stringValue   String valueOf integerValue         catch  NumberFormatException ex          JOptionPane showMessageDialog frame   Invalid input string         System out println  Invalid input string         return     Option  2  Reset the affected variable if the execution flow can continue in case of an exception  For example  with some modifications in the catch block  catch  NumberFormatException ex        integerValue   0       Using a string constant for comparison or any sort of computing is always a good idea  because a constant never returns a null value

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Converting a string to an int is more complicated than just converting a number  You have think about the following issues   Does the string only contain numbers 0-9  What s up with -   before or after the string  Is that possible  referring to accounting numbers   What s up with MAX - MIN INFINITY  What will happen if the string is 99999999999999999999  Can the machine treat this string as an int

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Convert a string to an integer with the parseInt method of the Java Integer class  The parseInt method is to convert the String to an int and throws a NumberFormatException if the string cannot be converted to an int type   Overlooking the exception it can throw  use this   int i   Integer parseInt myString     If the String signified by the variable myString is a valid integer like    1234        200        1     and it will be converted to a Java int  If it fails for any reason  the change can throw a NumberFormatException  so the code should be a little longer to account for this   For example  Java String to int conversion method  control for a possible NumberFormatException  public class JavaStringToIntExample     public static void main  String   args             String s    test       Use this if you want to test the exception below     String s    1234        try                The String to int conversion happens here       int i   Integer parseInt s trim               Print out the value after the conversion       System out println  int i       i             catch  NumberFormatException nfe              System out println  NumberFormatException      nfe getMessage                   If the change attempt fails     in this case  if you can try to convert the Java String test to an int     the Integer parseInt process will throw a NumberFormatException  which you must handle in a try catch block

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There are different ways of converting a string int value into an Integer data type value  You need to handle NumberFormatException for the string value issue   Integer parseInt  foo   Integer parseInt myString     Integer valueOf  foo   Integer valueOf myString     Using Java 8 Optional API  foo   Optional of myString  map Integer  parseInt  get

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Do it manually  public static int strToInt String str       int i   0      int num   0      boolean isNeg   false          Check for negative sign  if it s there  set the isNeg flag     if  str charAt 0      -             isNeg   true          i   1                Process each character of the string      while  i  lt  str length              num    10          num    str charAt i    -  0      Minus the ASCII code of  0  to get the value of the charAt i                if  isNeg          num   -num      return num

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Well  a very important point to consider is that the Integer parser throws NumberFormatException as stated in Javadoc    int foo  String StringThatCouldBeANumberOrNot    26263Hello     will throw exception String StringThatCouldBeANumberOrNot2    26263     will not throw exception try         foo   Integer parseInt StringThatCouldBeANumberOrNot     catch  NumberFormatException e            Will Throw exception          do something  anything to handle the exception     try         foo   Integer parseInt StringThatCouldBeANumberOrNot2     catch  NumberFormatException e            No problem this time  but still it is good practice to care about exceptions          Never trust user input            Do something  Anything to handle the exception      It is important to handle this exception when trying to get integer values from split arguments or dynamically parsing something

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Use Integer parseInt   and put it inside a try   catch block to handle any errors just in case a non-numeric character is entered  for example   private void ConvertToInt        String string   txtString getText        try          int integerValue Integer parseInt string           System out println integerValue             catch Exception e          JOptionPane showMessageDialog            Error converting string to integer n    e toString            Error            JOptionPane ERROR MESSAGE

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An alternate solution is to use Apache Commons  NumberUtils   int num   NumberUtils toInt  1234      The Apache utility is nice because if the string is an invalid number format then 0 is always returned  Hence saving you the try catch block   Apache NumberUtils API Version 3 4

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For a normal string you can use   int number   Integer parseInt  1234      For a String builder and String buffer you can use   Integer parseInt myBuilderOrBuffer toString

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String myString    quot 1234 quot   int foo   Integer parseInt myString    If you look at the Java documentation you ll notice the  quot catch quot  is that this function can throw a NumberFormatException  which of course you have to handle  int foo  try      foo   Integer parseInt myString     catch  NumberFormatException e       foo   0      This treatment defaults a malformed number to 0  but you can do something else if you like   Alternatively  you can use an Ints method from the Guava library  which in combination with Java 8 s Optional  makes for a powerful and concise way to convert a string into an int  import com google common primitives Ints   int foo   Optional ofNullable myString    map Ints  tryParse    orElse 0

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Simply you can try this   Use Integer parseInt your string   to convert a String to int Use Double parseDouble your string   to convert a String to double  Example String str    quot 8955 quot   int q   Integer parseInt str   System out println  quot Output gt  gt  gt   quot    q      Output  8955   String str    quot 89 55 quot   double q   Double parseDouble str   System out println  quot Output gt  gt  gt   quot    q      Output  89 55

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Try this code with different inputs of String   String a    10     String a    10ssda     String a   null   String a    12102    if null    a        try           int x   Integer ParseInt a trim              Integer y   Integer valueOf a trim                 It will throw a NumberFormatException in case of invalid string like   10ssda  or  123 212   so  put this code into try catch       catch NumberFormatException ex               ex getMessage

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you can use their wayes  String stringNumber    quot 0123456789 quot    1- Use parseInt try      int number   Integer parseInt stringNumber    catch  NumberFormatException e         Handel exception  mybe has space or alphabet     2- Use valueOf try      int number   Integer valueOf stringNumber    catch  NumberFormatException e         Handel exception  mybe has space or alphabet

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We can use the parseInt String str  method of the Integer wrapper class for converting a String value to an integer value   For example   String strValue    12345   Integer intValue   Integer parseInt strVal     The Integer class also provides the valueOf String str  method   String strValue    12345   Integer intValue   Integer valueOf strValue     We can also use toInt String strValue  of NumberUtils Utility Class for the conversion   String strValue    12345   Integer intValue   NumberUtils toInt strValue

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Here we go String str    quot 1234 quot   int number   Integer parseInt str   print number     1234

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import java util     public class strToint        public static void main String   args             String str    123           byte barr     str getBytes             System out println Arrays toString barr            int result   0           for int i   0  i  lt  barr length  i                    System out print barr i                    int ii   barr i               char a    char  ii              int no   Character getNumericValue a               result   result   10   no              System out println result                      System out println  result   result

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Apart from all these answer i found one new way  although it uses internally Integer parseInt    By using import javafx util converter IntegerStringConverter   new IntegerStringConverter   fromString  quot 1234 quot   intValue    or new IntegerStringConverter   fromString  quot 1234 quot    Although it is a little bit costly as new objects get created  I just wanted to add as I learned a new way  Just go through javafx util StringConverter lt T gt  class it helps to convert any wrapper class value to a string or vice versa

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I wrote this fast method to parse a string input into int or long  It is faster than the current JDK 11 Integer parseInt or Long parseLong  Although  you only asked for int  I also included the long parser  The code parser below requires that the parser s method must be small for it to operate quickly  An alternative version is below the test code  The alternative version is pretty quick and it does not depend on the size of the class   This class checks for overflow  and you could customize the code to adapt to your needs  An empty string will yield 0 with my method but that is intentional  You can change that to adapt your case or use as is   This is only the part of the class where parseInt and parseLong are needed  Note that this only deals with base 10 numbers   The test code for the int parser is below the code below         Copyright 2019 Khang Hoang Nguyen    Permission is hereby granted  free of charge  to any person obtaining a copy of this software and associated documentation files  the  Software    to deal in the Software without restriction  including without limitation the rights to use  copy  modify  merge  publish  distribute  sublicense  and or sell copies of the Software  and to permit persons to whom the Software is furnished to do so  subject to the following conditions    The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software     THE SOFTWARE IS PROVIDED  AS IS   WITHOUT WARRANTY OF ANY KIND  EXPRESS OR IMPLIED  INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY  FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT  IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM  DAMAGES OR OTHER LIABILITY  WHETHER IN AN ACTION OF CONTRACT  TORT OR OTHERWISE  ARISING FROM  OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE      author  Khang Hoang Nguyen - kevin fai host       final class faiNumber      private static final long   longpow    0L  1L  10L  100L  1000L  10000L  100000L  1000000L  10000000L  100000000L  1000000000L                                             10000000000L  100000000000L  1000000000000L  10000000000000L  100000000000000L                                             1000000000000000L  10000000000000000L  100000000000000000L  1000000000000000000L                                                    private static final int   intpow     0  1  10  100  1000  10000                                            100000  1000000  10000000  100000000  1000000000                                                            parseLong String str  parse a String into Long         All errors throw by this method is NumberFormatException         Better errors can be made to tailor to each use case               public static long parseLong final String str            final int length   str length            if  length    0              return 0L           char c1   str charAt 0           int start           if  c1     -     c1                       if  length    1                  throw new NumberFormatException String format  Not a valid long value  Input   s     str                start   1            else               start   0                                   Note  if length  gt  19  possible scenario is to run through the string            to check whether the string contains only valid digits             If the check had only valid digits then a negative sign meant underflow  else  overflow                      if  length - start  gt  19              throw new NumberFormatException String format  Not a valid long value  Input   s     str             long c          long out   0L           for     start  lt  length  start                  c    str charAt start     0                if  c  gt  9L                  throw new NumberFormatException  String format  Not a valid long value  Input   s     str                 out    c   longpow length - start                      if  c1     -                 out    out   1L                 If out  gt  0 number underflow supposed to be negative               if  out  gt  0L                  throw new NumberFormatException String format  Not a valid long value  Input   s     str                return out                       If out  lt  0 number overflow  supposed to be positive           if  out  lt  0L              throw new NumberFormatException String format  Not a valid long value  Input   s     str            return out                        parseInt String str  parse a string into an int         return 0 if string is empty               public static int parseInt final String str            final int length   str length            if  length    0              return 0           char c1   str charAt 0           int start           if  c1     -     c1                       if  length    1                  throw new NumberFormatException String format  Not a valid integer value  Input   s     str                start   1            else               start   0                     int out   0  int c          int runlen   length - start           if  runlen  gt  9                if  runlen  gt  10                  throw new NumberFormatException String format  Not a valid integer value  Input    s     str                 c    str charAt start     0        lt - Any number from 0 - 255   48 will yield greater than 9 except 48 - 57             if  c  gt  9                  throw new NumberFormatException String format  Not a valid integer value  Input    s     str                if  c  gt  2                  throw new NumberFormatException String format  Not a valid integer value  Input    s     str                out    c   intpow length - start                        for     start  lt  length  start                  c    str charAt start     0                if  c  gt  9                  throw new NumberFormatException String format  Not a valid integer value  Input    s     str                out    c   intpow length - start                      if  c1     -                 out    out   1              if  out  gt  0                  throw new NumberFormatException String format  Not a valid integer value  Input    s     str                return out                     if  out  lt  0              throw new NumberFormatException String format  Not a valid integer value  Input    s     str            return out            Test Code Section  This should take around 200 seconds or so      Int Number Parser Test  long start   System currentTimeMillis    System out println  INT PARSER TEST    for  int i   Integer MIN VALUE  i    Integer MAX VALUE  i        if  faiNumber parseInt    i     i         System out println  Wrong       if  i    0         System out println  HalfWay Done       if  faiNumber parseInt      Integer MAX VALUE     Integer MAX VALUE      System out println  Wrong    long end   System currentTimeMillis    long result    end - start   System out println result      INT PARSER END      An alternative method which is also very fast  Note that array of int pow is not used  but a mathematical optimization of multiply by 10 by bit shifting   public static int parseInt final String str        final int length   str length        if  length    0          return 0       char c1   str charAt 0       int start       if  c1     -     c1                   if  length    1              throw new NumberFormatException String format  Not a valid integer value  Input   s     str            start   1        else           start   0             int out   0      int c      while  start  lt  length  amp  amp  str charAt start      0           start        lt -- This to disregard leading 0  It can be                     removed if you know exactly your source                     does not have leading zeroes      int runlen   length - start       if  runlen  gt  9            if  runlen  gt  10              throw new NumberFormatException String format  Not a valid integer value  Input    s     str             c    str charAt start       0         lt - Any number from 0 - 255   48 will yield greater than 9 except 48 - 57         if  c  gt  9              throw new NumberFormatException String format  Not a valid integer value  Input    s     str            if  c  gt  2              throw new NumberFormatException String format  Not a valid integer value  Input    s     str            out    out  lt  lt  1     out  lt  lt  3    c      lt - Alternatively this can just be out   c or c above can just be out             for     start  lt  length  start              c    str charAt start     0            if  c  gt  9              throw new NumberFormatException String format  Not a valid integer value  Input    s     str            out    out  lt  lt  1     out  lt  lt  3    c             if  c1     -             out    out   1          if  out  gt  0              throw new NumberFormatException String format  Not a valid integer value  Input    s     str            return out             if  out  lt  0          throw new NumberFormatException String format  Not a valid integer value  Input    s     str        return out

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I am a little bit surprised that nobody mentioned the Integer constructor that takes String as a parameter  So  here it is  String myString    quot 1234 quot   int i1   new Integer myString    Java 8 - Integer String   Of course  the constructor will return type Integer  and an unboxing operation converts the value to int   Note 1  It s important to mention  This constructor calls the parseInt method  public Integer String var1  throws NumberFormatException       this value   parseInt var1  10      Note 2  It s deprecated   Deprecated since  quot 9 quot   - JavaDoc

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int foo   Integer parseInt  quot 1234 quot     Make sure there is no non-numeric data in the string

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Whenever there is the slightest possibility that the given String does not contain an Integer  you have to handle this special case  Sadly  the standard Java methods Integer  parseInt and Integer  valueOf throw a NumberFormatException to signal this special case  Thus  you have to use exceptions for flow control  which is generally considered bad coding style   In my opinion  this special case should be handled by returning an empty Optional lt Integer gt   Since Java does not offer such a method  I use the following wrapper   private Optional lt Integer gt  tryParseInteger String string        try           return Optional of Integer valueOf string          catch  NumberFormatException e            return Optional empty              Example usage      prints  12  System out println tryParseInteger  12   map i - gt  i toString    orElse  invalid        prints  -1  System out println tryParseInteger  -1   map i - gt  i toString    orElse  invalid        prints  invalid  System out println tryParseInteger  ab   map i - gt  i toString    orElse  invalid       While this is still using exceptions for flow control internally  the usage code becomes very clean  Also  you can clearly distinguish the case where -1 is parsed as a valid value and the case where an invalid String could not be parsed

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Alternatively  you can use Integer valueOf    It will return an Integer object   String numberStringFormat    10   Integer resultIntFormat   Integer valueOf numberStringFormat   LOG info  result     resultIntFormat     Output  10

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You could use any of the following   Integer parseInt s  Integer parseInt s  radix  Integer parseInt s  beginIndex  endIndex  radix  Integer parseUnsignedInt s  Integer parseUnsignedInt s  radix  Integer parseUnsignedInt s  beginIndex  endIndex  radix   Integer valueOf s  Integer valueOf s  radix  Integer decode s  NumberUtils toInt s  NumberUtils toInt s  defaultValue

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You can also begin by removing all non-numerical characters and then parsing the integer   String mystr   mystr replaceAll      d         int number   Integer parseInt mystr     But be warned that this only works for non-negative numbers

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As mentioned  Apache Commons  NumberUtils can do it  It returns 0 if it cannot convert a string to an int   You can also define your own default value   NumberUtils toInt String str  int defaultValue    Example   NumberUtils toInt  3244   1    3244 NumberUtils toInt     1        1 NumberUtils toInt null  5      5 NumberUtils toInt  Hi   6      6 NumberUtils toInt   32    1    1    Space in numbers are not allowed NumberUtils toInt StringUtils trimToEmpty    32    1     32

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One method is parseInt String   It returns a primitive int   String number    10   int result   Integer parseInt number   System out println result     The second method is valueOf String   and it returns a new Integer   object   String number    10   Integer result   Integer valueOf number   System out println result

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Custom algorithm   public static int toInt String value      int output   0    boolean isFirstCharacter   true    boolean isNegativeNumber   false    byte bytes     value getBytes      for  int i   0  i  lt  bytes length  i          char c    char  bytes i       if   Character isDigit c           isNegativeNumber    c     -          if    isFirstCharacter  amp  amp   isNegativeNumber    c                     throw new NumberFormatException  For input string       value                        else         int number   Character getNumericValue c         output   output   10   number            isFirstCharacter   false        if  isNegativeNumber      output    -1    return output      Another solution    Use the string charAt method instead of convert string to byte array   public static int toInt String value      int output   0    boolean isFirstCharacter   true    boolean isNegativeNumber   false    for  int i   0  i  lt  value length    i          char c   value charAt i       if   Character isDigit c           isNegativeNumber    c     -          if    isFirstCharacter  amp  amp   isNegativeNumber    c                     throw new NumberFormatException  For input string       value                        else         int number   Character getNumericValue c         output   output   10   number            isFirstCharacter   false        if  isNegativeNumber      output    -1    return output      Examples   int number1   toInt  20    int number2   toInt  -20    int number3   toInt   20    System out println  Numbers       number1          number2          number3    try     toInt  20 Hadi      catch  NumberFormatException e      System out println  Error      e getMessage

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I m have a solution  but I do not know how effective it is  But it works well  and I think you could improve it  On the other hand  I did a couple of tests with JUnit which step correctly  I attached the function and testing   static public Integer str2Int String str        Integer result   null      if  null    str    0    str length              return null            try           result   Integer parseInt str              catch  NumberFormatException e            String negativeMode               if str indexOf  -      -1              negativeMode    -           str   str replaceAll  -                 if  str indexOf         -1                str   str substring 0  str indexOf                    if  str length      0                    return  Integer 0                                  String strNum   str replaceAll      d                  if  0    strNum length                  return null                    result   Integer parseInt negativeMode   strNum             return result      Testing with JUnit    Test public void testStr2Int         assertEquals  is numeric    Integer  -5   Helper str2Int  -5         assertEquals  is numeric    Integer 50  Helper str2Int  50 00         assertEquals  is numeric    Integer 20  Helper str2Int    20 90         assertEquals  is numeric    Integer 5  Helper str2Int   5 321         assertEquals  is numeric    Integer 1000  Helper str2Int  1 000 50         assertEquals  is numeric    Integer 0  Helper str2Int  0 50         assertEquals  is numeric    Integer 0  Helper str2Int   50         assertEquals  is numeric    Integer 0  Helper str2Int  - 10         assertEquals  is numeric    Integer Integer MAX VALUE  Helper str2Int    Integer MAX VALUE        assertEquals  is numeric    Integer Integer MIN VALUE  Helper str2Int    Integer MIN VALUE        assertEquals  Not      is numeric   null  Helper str2Int  czv  xcvsa                    Dynamic test             for Integer num   0  num  lt  1000  num              for int spaces   1  spaces  lt  6  spaces                  String numStr   String format   0  spaces  d   num               Integer numNeg   num   -1              assertEquals numStr      is numeric   num  Helper str2Int numStr                assertEquals numNeg      is numeric   numNeg  Helper str2Int  -     numStr

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This can work  Integer parseInt yourString

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It can be done in seven ways  import com google common primitives Ints  import org apache commons lang math NumberUtils   String number    quot 999 quot     Ints tryParse  int result   Ints tryParse number    NumberUtils createInteger  Integer result   NumberUtils createInteger number    NumberUtils toInt  int result   NumberUtils toInt number    Integer valueOf  Integer result   Integer valueOf number    Integer parseInt  int result   Integer parseInt number    Integer decode  int result   Integer decode number    Integer parseUnsignedInt  int result   Integer parseUnsignedInt number

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Apart from the previous answers  I would like to add several functions  These are results while you use them   public static void main String   args      System out println parseIntOrDefault  123   0       123   System out println parseIntOrDefault  aaa   0       0   System out println parseIntOrDefault  aaa456   3  0       456   System out println parseIntOrDefault  aaa789bbb   3  6  0       789     Implementation   public static int parseIntOrDefault String value  int defaultValue      int result   defaultValue    try       result   Integer parseInt value         catch  Exception e          return result     public static int parseIntOrDefault String value  int beginIndex  int defaultValue      int result   defaultValue    try       String stringValue   value substring beginIndex       result   Integer parseInt stringValue         catch  Exception e          return result     public static int parseIntOrDefault String value  int beginIndex  int endIndex  int defaultValue      int result   defaultValue    try       String stringValue   value substring beginIndex  endIndex       result   Integer parseInt stringValue         catch  Exception e          return result

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You can use new Scanner  1244   nextInt    Or ask if even an int exists  new Scanner  1244   hasNextInt

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Google Guava has tryParse String   which returns null if the string couldn t be parsed  for example   Integer fooInt   Ints tryParse fooString   if  fooInt    null

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Methods to do that   Integer parseInt s  Integer parseInt s  radix  Integer parseInt s  beginIndex  endIndex  radix  Integer parseUnsignedInt s  Integer parseUnsignedInt s  radix  Integer parseUnsignedInt s  beginIndex  endIndex  radix  Integer valueOf s  Integer valueOf s  radix  Integer decode s  NumberUtils toInt s  NumberUtils toInt s  defaultValue   Integer valueOf produces an Integer object and all other methods a primitive int  The last two methods are from commons-lang3 and a big article about converting here

[java] How do I convert a String to an int in Java?

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