While casting from Char to int, I am getting ASCII value of the Char. How do I get the int value?-CodePudding

public static void main(String[] args) {
    Scanner scnr = new Scanner(System.in);
    String user_input = scnr.nextLine();
    System.out.println(user_input.charAt(0));
    int listSize = (int) (user_input.charAt(0));
    String[] wordsList = new String[listSize];
    user_input = user_input.substring(2);
    wordsList = user_input.split(" ");
    for (int i =0; i<listSize; i  ){
        int frequency = getWordFrequency(wordsList, listSize, wordsList[i]);
        System.out.println(wordsList[i]  " "   frequency );
    }


}

This is the code if I give "5 hey Hi Mark hi mark" the list size is the ASCII value of 5 instead of int 5.

Tried changing the char directly into the int value. It is giving me an ASCII value.

CodePudding user response：

There are more characters representing digits than just 0 to 9. For example, there's ۲, which is the arabic-indic variant of 2.

Hence, use the right unicode aware methods for the job:

int value = Character.digit('\u06F2', 10);
System.out.println(value);

This will print '2'. 06F2 is the unicode point for ۲, you can just put ۲ straight in the source file, but if your compiler and editor don't agree on charset encoding it would fail, the backslash u escape is 'safer'.

If you're limited to '0'-'9', they take up consecutive spots in the unicode table, so you can subtract '0':

char v = '5';
int c = v - '0';
System.out.println(c); // prints 5

But, don't. Use Character.digit. The , 10 in the digit method is the 'radix', you probably want 10. Feel free to peruse the javadoc of Character::digit.

If you have string containing a number (consisting of potentially more than one digit symbol), theres Integer.parseInt, which is also capable of parsing most number symbols:

String value = "\u06F2\u06F2";
System.out.println(Integer.parseInt(value));

prints 22.

CodePudding user response：

Subtract '0' or use Character.getNumericValue.

int listSize = user_input.charAt(0) - '0';

For integers with more than one digit, use Integer.parseInt.