How to convert a String to int in Java?

Query:

My String contains only numbers, and I want to return the number it represents. For example, given the string "1234" the result should be the number 1234.

In this post, I’ll convert how to convert Integer to String in Java as well.

How to convert a String to int in Java?

In Java you can convert String to Integer by the following methods:

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

If you look at the Java documentation you’ll notice the “catch” is that this function can throw a NumberFormatException, which of course you have to handle:

int foo;
try {
   foo = Integer.parseInt(myString);
}
catch (NumberFormatException e)
{
   foo = 0;
}

(This treatment defaults a malformed number to 0, but you can do something else if you like.)

Alternatively, you can use an Ints method from the Guava library, which in combination with Java 8’s Optional, makes for a powerful and concise way to convert a string into an int:

import com.google.common.primitives.Ints;

int foo = Optional.ofNullable(myString)
 .map(Ints::tryParse)
 .orElse(0)

Methods to convert String to integer in Java:

For example, here are two ways:

Convert String to integer using valueOf():

Integer x = Integer.valueOf(str);

Convert String to Integer using parseInt():

int y = Integer.parseInt(str);

There is a slight difference between parseInt() and valueOf() methods:

  • valueOf returns a new or cached instance of java.lang.Integer
  • parseInt returns primitive int.

The same is for all cases: Short.valueOf/parseShortLong.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.

Answer #3:

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 cycles, therefore, a for cycle (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.

We’re dividing the latter (lastdigit) by 10 in the fashion 2/10 = 0.2 (hence why double) like this:

 lastdigit = lastdigit / 10;

This is merely playing with numbers. We were 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 to 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 cycles
  • No “Magic” Expressions such as parseInt

Answer #4:

Apart from the previous answers, I would like to add several functions. These are results while you use 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
}

Implementation:

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;
}

How to convert Integer to String in Java?

There are multiple ways:

  • String.valueOf(number) (my preference)
  • "" + number (I don’t know how the compiler handles it, perhaps it is as efficient as the above)
  • Integer.toString(number)

Integer class has static method toString() – you can use it:

int i = 1234;
String str = Integer.toString(i);

Returns a String object representing the specified integer. The argument is converted to signed decimal representation and returned as a string, exactly as if the argument and radix 10 were given as arguments to the toString(int, int) method.

Always use either String.valueOf(number) or Integer.toString(number).

Using “” + number is an overhead and does the following:

StringBuilder sb = new StringBuilder();
sb.append("");
sb.append(number);
return sb.toString();

Answer #2:

The way I know how to convert an integer into a string is by using the following code:

Integer.toString(int);

and

String.valueOf(int);

If you had an integer i, and a string s, then the following would apply:

int i;
String s = Integer.toString(i); or
String s = String.valueOf(i);

If you wanted to convert a string “s” into an integer “i”, then the following would work:

i = Integer.valueOf(s).intValue();

In this post, I shared answers to how to convert String to Integer and vice-versa in Java.

Hope you learned something from this post.

Follow Programming Articles for more!

About ᴾᴿᴼᵍʳᵃᵐᵐᵉʳ

Linux and Python enthusiast, in love with open source since 2014, Writer at programming-articles.com, India.

View all posts by ᴾᴿᴼᵍʳᵃᵐᵐᵉʳ →