How do I convert a numeric string input from a user into an integer array in Java?
What I did so far is:
# Get User input
System.out.println("Guess the number: ");
String input = read.nextLine();
# String to int
int digitNumber = Integer.parseInt(input);
I'm trying to convert a numeric string (e.g., "089") entered by a user into an integer array where each digit becomes an individual element. For example, "089" should result in [0, 8, 9]. However, using Integer.parseInt(input) just gives me 89 and drops the leading zero.
What would be the best way to convert this input into an array of integers?
int[] numbers =
input
.codePoints() // Generate a stream of code point `int` numbers for the characters in this string.
.filter ( Character :: isDigit ) // Ignore any non-digit.
.mapToObj ( Character :: toString ) // Convert each `int` code point to its character. We expect that character to be a digit.
.mapToInt ( Integer :: parseInt ) // Parse the textual digit as an `int` primitive value.
.toArray (); // Collect those `int` primitive values into an array, `int[]`.
charThe Answer by Stultuske is correct, except that you should avoid using char type.
The char type has been essentially broken since Java 2, and legacy since Java 5. As a 16-bit value, char is physically incapable of representing most characters.
Make a habit of using integer code point numbers rather than char values.
Calling String#codePoints generates an IntStream of int primitive values, one int for each character in that string successively.
String input = "089" ;
int[] codePoints = input.codePoints().toArray() ;
int[] numbers = new int[ codePoints.length ] ;
int index = 0 ;
for ( int codePoint : codePoints )
{
String digit = Character.toString ( codePoint ) ;
int number = Integer.parseInt ( digit ) ;
numbers [ index ] = number ;
index = index + 1 ;
}
System.out.println ( Arrays.toString( numbers ) ) ;
See this code run at Ideone.com.
[0, 8, 9]
As Rob Spoor reminds us in a Comment, we should verify that each input code point does indeed represent a digit.
You could check for just the Latin 0-9 digits, or more expansively check for any script’s digits.
String input = "0Z9" ; // Simplate FAULTY INPUT.
int[] codePoints = input.codePoints().toArray() ;
int[] numbers = new int[ codePoints.length ] ;
int index = 0 ;
for ( int codePoint : codePoints )
{
String digit = Character.toString ( codePoint ) ;
if ( Character.isDigit ( codePoint ) ) // Verify the code point does indeed represent a character that is a digit.
{
int number = Integer.parseInt ( digit ) ;
numbers [ index ] = number ;
index = index + 1 ;
}
else
{
System.out.println ( "ERROR - Received non-digt. Code point: " + codePoint + ". Character: " + digit ) ;
}
}
System.out.println ( Arrays.toString( numbers ) ) ;
See this code run at Ideone.com.
Notice this throws off our assumption in sizing our results array.
ERROR - Received non-digt. Code point: 90. Character: Z
[0, 9, 0]
Using a List rather than an array would help remedy that issue. And using modern Java we can further simplify this code.
String input = "0Z89" ; // Simulate FAULTY INPUT.
int[] numbers =
input
.codePoints() // Generate a stream of code point `int` numbers for the characters in this string.
.filter ( Character :: isDigit ) // Ignore any non-digit.
.mapToObj ( Character :: toString ) // Convert each `int` code point to its character. We expect that character to be a digit.
.mapToInt ( Integer :: parseInt ) // Parse the textual digit as an `int` primitive value.
.toArray (); // Collect those `int` primitive values into an array, `int[]`.
System.out.println ( Arrays.toString( numbers ) ) ;
When run at Ideone.com:
[0, 8, 9]