How can i convert a string to a abstract number, provided string is any valid number in java (say int, long, double etc).
I will not know the type of number in the string, so i can't use specific primitive parsing (like Integer.parseInt, Long.parseLong etc). Is there any generic way to convert it?
Eg:
Use NumberFormat. Number cannot be instantiated because it is an abstract class.
Number number = NumberFormat.getInstance().parse(string);
NumberFormat.getInstance Returns a general-purpose number format for the current default FORMAT locale. Be aware of that if you don't want to parse 2.000 as either 2 or 2000 depending on locale - you should specify a locale that matches the expected format of your input.Two cases:
double primitiveNumber = Double.parseDouble(input);
Double will make your value lose precision if too long.
You should use a BigDecimal instead:
BigDecimal number = new BigDecimal("43.256");
You can then get different primitives like this:
try {
int intValue = number.intValueExact();
} catch (ArithmeticException e) {
try {
long longValue = number.longValueExact();
} catch (ArithmeticException e) {
double doubleValue = number.doubleValue();
}
}
Double.valueOf() will do fine unless you have a very long number.
valueOf() returns a Double, not a primitive.
The quickest way to do this is just always use Double.parseDouble(). If you need a general-purpose parser that determines which primitive to use, rather than always using the same type, then you will need to write your own. One way to do this is to use each parseXxx() method and lot of try...catch statements.
For integers:
int myInteger = Integer.parseInt("23423");
For doubles:
double myDouble = Double.parseDouble("34.3");
As long as your strings could be big numbers (suppose that no longer than primitive long and double), then generic way could be decide whether it is long or double.
Number n = Double.valueOf(string);
if((double)n.intValue() == n.doubleValue()){
//n is double use n.doubleValue() or just use the n for generic purposes
}
else{
//n is int use n.intValue() or just use the n for generic purposes
}
I do prefer BigDecimal only. Because it won't have the issues with size and precision
You can use Double.parseDouble() to convert your string to double. This will work even if the string is an integer:
String myString = "1234.56";
double myDouble = Double.parseDouble(myString);
If you need to check if it is an integer or a double:
String myString = "1234.56";
int myInt;
double myDouble = Double.parseDouble(myString);
if ((myDouble % 1) == 0) {
myInt = (int) myDouble;
System.out.println("myString is an integer: " + myInt );
} else {
System.out.println("myString is a double: " + myDouble );
}
Or you can use Guava's DoubleMath.isMathematicalInteger():
String myString = "1234.56";
int myInt;
double myDouble = Double.parseDouble(myString);
if (DoubleMath.isMathematicalInteger(myDouble)) {
myInt = (int) myDouble;
System.out.println("myString is an integer: " + myInt );
} else {
System.out.println("myString is a double: " + myDouble );
}
Double.parseDouble(). This will work for all of the above examples.Numberclass ?