convert string to integer java without integer.parseint

Convert string to integer java without integer.parseint code#

Convert string to integer java without integer.parseint code#

I copied the code from rally25rs answer and added some tests for non-integer data. Which opposed to the previous benchmarks, this one shows Regex solution actually has consistently good performance.Ĭopied from Bill the Lizard and updated with compiled version: private final Pattern pattern = pile('^-?\d+$') So i'm gonna give an updated version of the benchmark, with a compiled version of the Regex. Since there's possibility that people still visit here and will be biased against Regex after the benchmarks. I do agree that Jonas K's solution is the most robust too. Long startTime = System.currentTimeMillis() īyRegex: 453 (note: re-compiling the pattern every time) When staying in the same method, they aren't bad performers. Exceptions aren't actually that expensivve, unless you start popping back multiple methods and the JVM has to do a lot of work to get the execution stack in place. However, autoboxing and auto-unboxing results in performance overhead, so it is preferable to rely on Integer.parseInt() to convert strings to the int type, and Integer.valueOf() to convert strings to Integer type.Did a quick benchmark. This means, you can technically write statements like these: Integer num1 = Integer.parseInt( '742') Įven though Integer.parseInt() returns a int, it can be assigned to a Integer variable, and in a similar manner, even though Integer.valueOf() returns a Integer, it can be assigned to an int.

To make programming easier, Java 8 and above transparently convert primitive types to their corresponding classes (such as int to Integer), and the other way round through autoboxing and auto-unboxing. 1010 in binary is 10 in decimal, so it prints 10 We can also provide an optional radix for base conversions: String str2 = '1010'

Just as with Integer.parseInt(), it will throw a NumberFormatException in case the number can't be converted: // 742a is not a valid number, so it prints 742a cannot be converted String str1 = '742a' this statement is WRONG as we cannot get the class of a primitive type getClass() on the int variable - it will result in a compile-time error. On the other hand, Integer.parseInt() returns the primitive type int, and we can confirm this by calling. ( 'num1 = ' + num1 + ' and it is of ' + num1.getClass()) In the program below, we convert a string to an Integer using Integer.valueOf() and print the class of the object: String str1 = '742' The Integer.valueOf() method is similar to Integer.parseInt() method, except it returns an Integer object rather than the primitive int type. Similar to how hexadecimal uses the numbers 0-9 and the letters a-f to represent a hexadecimal digit, base-36 uses the numbers 0-9 and the letters a-z to represent numbers. The number ad in hexadecimal can be converted to decimal as a (10) × 16 1 + d (13) × 16 0 = 173, and we get this value as the output.

In the second case, we're converting str2 to an int in base-16, aka, the hexadecimal number system.

The number 1010 in binary can be converted to decimal as 1 × 2 3 + 0 × 2 2 + 1 × 2 1 + 0 × 2 0 = 10, and we get the same as the output.

In the first case, we're converting str1 to an int in base-2, aka, the binary number system.

This is useful when you want to convert a string containing a number in a different base (such as binary, octal or hexadecimal), and it can be used like so: String str1 = '1010' (str2 + ' cannot be converted') įinally, you can also provide an optional radix to the Integer.parseInt() method. as this string is too long, it will print 2222222222 cannot be converted String str2 = '2222222222' An example is shown below: // as this string has a letter, it will print 742a cannot be converted String str1 = '742a' You can handle this exception using try/catch and show an error message, or perform any other suitable action. If the string contains letters, has invalid symbols, or is simply too large to fit inside the range of an int (-2147483648 to 2147483647), the method throws an exception of the type NumberFormatException. An example is shown below: String str1 = '742' The Integer.parseInt() method in Java takes in a string and returns an int representing the number in the string. Converting a string to an integer is a common task and Java provides a number of methods for this purpose. Java is a versatile language that can be used for a variety of purposes.