Converting String to Int in Java?

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.

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.

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 < 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;
}

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 = "12";

Another limitation was the fact that I couldn't use repetitive cicles, therefore, a for cicle (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 "weird" 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. What we're doing is divide the latter (lastdigit) by 10, in this fashion 2/10 = 0.2 (hence why double) like this:

 lastdigit = lastdigit/10;

This is merely playing with numbers. What we did here was 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 "box" 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 "()"

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 Cicles
• No "Magic" Expressions such as parseInt

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

Converting a string to an int is more complicated than just convertig a number. You have think about the following issues:

• Does the string only contains 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?

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 < 1000; num++) {
        for(int spaces = 1; spaces < 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));
        }
    }
}

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);

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:

1. Unboxing:

   int val = Integer.decode("12"); 
   

2. intValue():

   Integer.decode("12").intValue();
   

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 Optional<Integer>. Since Java does not offer such a method, I use the following wrapper:

private Optional<Integer> tryParseInteger(String string) {
    try {
        return Optional.of(Integer.valueOf(string));
    } catch (NumberFormatException e) {
        return Optional.empty();
    }
}

Usage:

// prints 1234
System.out.println(tryParseInteger("1234").orElse(-1));
// prints -1
System.out.println(tryParseInteger("foobar").orElse(-1));

While this is still using exceptions for flow control internally, the usage code becomes very clean.

You can also begin by removing all non numerical characters and then parsing the int:

string mystr = mystr.replaceAll( "[^\\d]", "" );
int number= Integer.parseInt(mystr);

But be warned that this only works for non negative numbers.

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.

Guava has tryParse(String), which returns null if the string couldn't be parsed, for example:

Integer fooInt = Ints.tryParse(fooString);
if (fooInt != null) {
  ...
}

As mentioned Apache Commons NumberUtils can do it. Which return 0 if it cannot convert string to 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; 

Apart from these above answers, I would like to add several functions:

    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;
  }

And here are results while you running 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
  }

For normal string you can use:

int number = Integer.parseInt("1234");

For String builder and String buffer you can use:

Integer.parseInt(myBuilderOrBuffer.toString());

Just for fun. You can use java 8 Optional for converting a String into anInteger.

String str = "123"; 
Integer value = Optional.of(str).map(Integer::valueOf).get();
// will return Integer value of specified String, or it
// will throw NPE when str is null.     
value = Optional.ofNullable(str).map(Integer::valueOf).orElse(-1);
// will do the same as the code above, except it will return -1 
// when srt is null, instead of throwing a NPE.

We just combine here Integer.valueOf and Optinal. Probably there might be situations when this is useful - for example when you want to avoid null checks. Pre java 8 code will look like this:

Integer value = (str == null) ? -1 : Integer.parseInt(str);

String str="1234";

int foo=Integer.parseInt("1234");

make sure there is no non-numeric data in the string.

For example, here are two ways:

   Integer x = Integer.valueOf(str);
    // or
   int y = Integer.parseInt(str);

String s="100";
try {
 int i=Integer.parseInt( s );
}catch( Exception e ) {
 System.out.println(e.getMessage());
}



String s="100L";
try {
  int i=Integer.parseInt( s );
 }catch( Exception e ) {
  System.out.println(e.getMessage());
}

you need to use parseInt for

int value = Integer.parseInt("123"); 

String st = "123";

int num = Integer.parseInt(st);