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

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

Gareth D.

The ProblemJump To Solution

How can I convert a String to an int in Java?

The Solution

The two easiest ways to convert a string to an integer in Java are to use Integer.parseInt() or Integer.valueOf().

Here is an example of each. Note that both could throw a NumberFormatException if the input string contains characters that cannot be converted to an integer, so your code should handle this in a try...catch block.

Using Integer.valueOf

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class Main { public static void main(String[] args) { String validString = "123"; String invalidString = "123x"; Integer number; try { number = Integer.valueOf(validString); System.out.println("Converted integer: " + number); number = Integer.valueOf(invalidString); System.out.println("Converted integer: " + number); } catch (NumberFormatException e) { System.out.println("Invalid integer input"); } } }

This will output the following.

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Converted integer: 123 Invalid integer input

In the above example, validString is successfully converted as it contains only numerical characters. The invalidString variable contains a non-numeric character and cannot be converted, so we instead see the output from the catch block.

Using Integer.parseInt()

The code below has only three differences from the previous example. We use int number; instead of Integer number to declare our variable that will hold the result, and we changed both instances of valueOf to parseInt. The output and other functionality are otherwise identical.

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class Main { public static void main(String[] args) { String validString = "123"; String invalidString = "123x"; int number; try { number = Integer.parseInt(validString); System.out.println("Converted integer: " + number); number = Integer.parseInt(invalidString); System.out.println("Converted integer: " + number); } catch (NumberFormatException e) { System.out.println("Invalid integer input"); } } }

Choosing Between Integer and int

Modern Java versions running on modern systems are very efficient and there should be practically no performance difference between using primitive int and declaring new Integer objects, so you should consider convenience over performance in nearly all cases. Although initializing new Integer objects will take up some extra memory, this will be trivial unless you are keeping vast amounts in memory simultaneously.

Instead, you might consider factors such as:

  • If you need to store the result in a database, then Integer is nullable which might be useful.
  • Integer includes other helper methods, which you might need if you do further processing after the conversion.
  • The primitive int can behave more predictably and manipulating int variables sometimes requires less code.

For example, consider the following:

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class Main { public static void main(String[] args) { String s1 = "1000"; String s2 = "1000"; Integer n1 = Integer.valueOf(s1); Integer n2 = Integer.valueOf(s2); System.out.println("n1 == n2: " + String.valueOf(n1 == n2)); } }

Even though both s1 and s2 represent the same number (1000), this prints out:

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n1 == n2: false

This is because n1 and n2 are different objects with different memory addresses.

Confusingly, the following code—which is identical except that it uses the value “100” instead of “1000”—behaves differently:

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class Main { public static void main(String[] args) { String s1 = "100"; String s2 = "100"; Integer n1 = Integer.valueOf(s1); Integer n2 = Integer.valueOf(s2); System.out.println("n1 == n2: " + String.valueOf(n1 == n2)); } }

This prints:

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n1 == n2: true

This difference in behavior is because Java caches smaller integer values for efficiency, so in the second example, both values are saved as the same object.

Using parseInt() and int primitives can avoid issues like this, giving you n1 == n2: true in both cases. However, using Integer has the benefit of the helper methods it contains and the ability to deal with null values.

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