Converting Strings to Numbers in Java

To convert a String object to a primitive type you can use the following methods offered by the wrapper classes.

Wrapper	Static parseXxx() method
Byte	parseByte(String s) parseByte(String s,int radix)
Short	parseShort(String s) parseShort(String s,int radix)
Integer	parseInt(String s) parseInt(String s,int radix)
Long	parseLong(String s) parseLong(String s,int radix)
Float	parseFloat(String s)
Double	parseDouble(String value)
Character	None
Boolean	parseBoolean(String s)
Void	None

For example :

        byte b1 = Byte.parseByte("10");
        byte b2 = Byte.parseByte("10", 16);

        short s1 = Short.parseShort("15");
        short s2 = Short.parseShort("15", 2);

        int i1 = Integer.parseInt("20");
        int i2 = Integer.parseInt("20", 8);

        long l1 = Long.parseLong("100");
        long l2 = Long.parseLong("100", 8);

        float f1 = Float.parseFloat("3.14");

        double f2 = Double.parseDouble("3.14");

        boolean bool = Boolean.parseBoolean("false");



Notice that Byte,Short ,Integer and Long parseXxx() methods take an additional argument, int radix, which indicates in what base (for example binary, octal, or hexadecimal) the first argument is represented.

for example,

int i1 = Integer.parseInt("1001",2);

Here converts binary 1001 to the corresponding decimal value 9 and assigns the value 9 to variable i1.

java.lang.NumberFormatException

 

The parseXxx() methods throw java.lang.NumberFormatException when the given input string is invalid.For example

int i1 = Integer.parseInt("20f");
The above line throws a run-time exception,  java.lang.NumberFormatException: For input string: "20f".