atoi in java using charAt

Question

I know that one solution to do this is the following:

 String tmp = "12345";
      int result = 0;
      for (int i =0; i < tmp.length(); i++){
          char digit = (char)(tmp.charAt(i) - '0');
          result += (digit * Math.pow(10, (tmp.length() - i - 1)));

      }

      System.out.println(result);

What I don't understand is why is:

char digit = (char)(tmp.charAt(i) - '0');

How can this convert into a digit?

Answer 1

char digit = (char)(tmp.charAt(i) - '0');

In the ascii table , characters from '0' to '9' are contiguous. So, if you know that tmp.charAt(i) will return a character between 0 and 9 , then subracting 0 will return the offset from zero, that is, the digit that that character represents.

Answer 2

Using Math.pow is very expensive, you would be better off using Integer.parseInt.

You don't have to use Math.pow. If your numbers are always positive you can do

int result = 0;
for (int i = 0; i < tmp.length(); i++)
   result = result * 10 + tmp.charAt(i) - '0';

Answer 3

char is an integer type that maps our letters to numbers a computer can understand (see an ascii chart ). A string is just an array of characters. Since the digits are contiguous in ascii representation, '1' - '0' = 49 - 48 = 1 , '2' - '0' = 50 - 48 = 2 , etc.

Answer 4

Try this:

int number = Integer.parseInt("12345") 
// or 
Integer number = Integer.valueOf("12345") 

atoi could be a bit mistery for developers. Java prefers more readable names

Answer 5

If using Integer.parseInt you have to catch exceptions, since Integer.parseInt("82.23") or Integer.parseInt("ABC") will launch exceptions.

If you want to allow things like

     atoi ("82.23") // => 82
     atoi ("AB123") // => 0

which makes sense then you can use

     public static int atoi (String sInt)
     {
        return (int) (atof (sInt));
     }

     public static long atol (String sLong)
     {
        return (long) (atof (sLong));
     }

     public static double atof (String sDob)
     {
        double reto = 0.;

        try {
           reto = Double.parseDouble(sDob);
        }
        catch (Exception e) {}

        return reto;
     }

Answer 6

If programming in Java, please use this,

int atoi(String str)
{
    try{
        return Integer.parseInt(str);
    }catch(NumberFormatException ex){
        return -1;   
    }
}

Answer 7

I echo what Tom said.

If you are confused with the above implementation then you can refer the below simpler implementation.

private static int getDecValue(char hex) {
    int dec = 0;
    switch (hex) {
    case '0':
        dec = 0;
        break;
    case '1':
        dec = 1;
        break;
    case '2':
        dec = 2;
        break;
    case '3':
        dec = 3;
        break;
    case '4':
        dec = 4;
        break;
    case '5':
        dec = 5;
        break;
    case '6':
        dec = 6;
        break;
    case '7':
        dec = 7;
        break;
    case '8':
        dec = 8;
        break;
    case '9':
        dec = 9;
        break;
    default:
        // do nothing
    }
    return dec;
}

public static int atoi(String ascii) throws Exception {
    int integer = 0;
    for (int index = 0; index < ascii.length(); index++) {
        if (ascii.charAt(index) >= '0' && ascii.charAt(index) <= '9') {
            integer = (integer * 10) + getDecValue(ascii.charAt(index));
        } else {
            throw new Exception("Is not an Integer : " + ascii.charAt(index));
        }
    }
    return integer;
}

Answer 8

Java has a built in function that does this...

String s =  "12345";
Integer result = Integer.parseInt(s);