how to convert a string to float and avoid using try/catch in java?

Question

There are some situation that I need to convert string to float or some other numerical data-type but there is a probability of getting some nonconvertible values such as "-" or "/" and I can't verify all the values beforehand to remove them. and I want to avoid using try/catch for this matter , is there any other way of doing a proper conversion in java? something similar to C# TryParse?

Answer 1

The simplest thing I can think of is java.util.Scanner . However this approach requires a new Scanner instance for each String.

String data = ...;
Scanner n = new Scanner(data);
if(n.hasNextInt()){//check if the next chars are integer
  int i = n.nextInt();   
}else{

}

Next you could write a regex pattern that you use to check the String (complex to fail too big values) and then call Integer.parseInt() after checking the string against it.

Pattern p = Pattern.compile("insert regex to test string here");
String data = ...;
Matcher m = p.matcher(data);
//warning depending on regex used this may 
//only check part of the string
if(m.matches()){
  int i = Integer.parseInt(data);
}

However both of these parse the string twice, once to test the string and a second time to get the value. Depending on how often you get invalid strings catching an exception may be faster.

Answer 2

Unfortunately, there is no such method in Java. There is no out parameter in Java, so writing such a method would need to return a null Float to signal an error, or to pass a FloatHolder object which could be modified by the method:

public class FloatHolder {
    private float value;

    public void setValue(float value) {
        this.value = value;
    }

    public float getValue() {
        return this.value;
    }
}

public static boolean tryParseFloat(String s, FloatHolder holder) {
    try {
        float value = Float.parseFloat(s);
        holder.setValue(value);
    }
    catch (NumberFormatException e) {
        return false;
    }
}

Answer 3

This is an old question, but since all the answers fail to mention this (and I wasn't aware of it myself until seeing it in a merge request written by a colleague), I want to point potential readers to the Guava Floats and Ints classes:

With the help of these classes, you can write code like this:

    Integer i = Ints.tryParse("10");
    Integer j = Ints.tryParse("invalid");
    Float f = Floats.tryParse("10.1");
    Float g = Floats.tryParse("invalid.value");

The result will be null if the value is an invalid int or float, and you can then handle it in any way you like. (Be careful to not just cast it to an int/float, since this will trigger a NullPointerException if the value is an invalid integer/floating point value.)

Note that these methods are marked as "beta", but they are quite useful anyway and we use them in production.

For reference, here are the Javadocs for these classes:

• https://google.github.io/guava/releases/snapshot-jre/api/docs/com/google/common/primitives/Ints.html
• https://google.github.io/guava/releases/snapshot-jre/api/docs/com/google/common/primitives/Floats.html

Answer 4

Java does not provide some built in tryParse type of methods, on of the solutions you can try is to create your own tryParse Method and put try/catch code in this method and then you can easily use this method across your application very easily and without using try/catch at all the places you use the method.

One of the sample functions can have following code

public static Long parseLong(String value) {

        if(isNullOrEmpty(value)) {
            return null;
        }

        try {
            return Long.valueOf(value);
        }
        catch (NumberFormatException e) {
        }

        return null;
    }

Answer 5

Regular expressions helped me solve this issue. Here is how:

1. Get the string input.
2. Use the expression that matches one or more digits.
3. Parse if it is a match.

String s = "1111";
int i = s.matches("^[0-9]+$") ? Integer.parseInt(s) : -1;
if(i != -1)
  System.out.println("Integer");
else
  System.out.println("Not an integer");