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?

yes I am.. and I'd have to write my own function, kind of an atoi in C — xonegirlz, Nov 07 '11 at 16:07
Is this homework? If not, just use `Integer.valueOf("12345")`. If it is, please tag it as such. — Ted Hopp, Nov 07 '11 at 16:08
@xonegirlz - did you try my solution? If it worked then please accept the answer and vote up! — JRG, Jul 26 '17 at 05:53
Possible duplicate of [How to convert a String to an int in Java?](https://stackoverflow.com/questions/5585779/how-to-convert-a-string-to-an-int-in-java) — Clijsters, Aug 15 '17 at 14:10
If @PTBG is right and you want to convert Strings to Ints, this question is a duplicate: https://stackoverflow.com/questions/5585779/how-to-convert-a-string-to-an-int-in-java — Clijsters, Aug 15 '17 at 14:11

score 13 · Answer 1 · answered Nov 07 '11 at 16:08

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.

score 11 · Answer 2 · answered Nov 07 '11 at 16:13

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';

score 7 · Answer 3 · answered Nov 07 '11 at 16:10

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.

lukastymo · Answer 4 · 2011-11-07T16:20:45.597

3

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

edited Nov 07 '11 at 16:20

answered Nov 07 '11 at 16:09

lukastymo

26,145
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elxala · Answer 5 · 2016-04-01T09:58:40.913

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;
     }

score 1 · Answer 6 · answered Aug 15 '17 at 13:43

1

If programming in Java, please use this,

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

answered Aug 15 '17 at 13:43

Bharathkumar V

327
5
13

1

While this is correct and - imho - answers the question, it qould benefit from more contextual information like, what `parseInt()` is, what it does, and where one can find it's documentation. – Clijsters Aug 15 '17 at 14:09
Sure. Static method `parseInt()` verifies if characters in string are decimals except for first one, which could be minus sign. [link]https://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#parseInt(java.lang.String) – Bharathkumar V Aug 15 '17 at 14:21
Please use the edit functionality on your question. See [How do I write a good answer](https://stackoverflow.com/help/how-to-answer) in the help center. – Clijsters Aug 15 '17 at 15:33

score 0 · Answer 7 · answered Jul 03 '17 at 22:51

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;
}

score 0 · Answer 8 · answered Nov 07 '11 at 16:08

0

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

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

answered Nov 07 '11 at 16:08

PTBG

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atoi in java using charAt

8 Answers8