[irmoathlete]

the goal is to take single integers and add them together of integers comprised of more than 1 integers.

for example, to take the list 1, 2 ,5 ,12 , 15, 8

and do the following:

1+2+5+(1+2) + (1+5) + 8

the numbers that need to be split have been added to an array list i just need to know how i can split single integers?

[Crunch]

Convert the individual integer to a string object

And then do a charAt operation to get the individual digit.

[bcranger]

If you follow Crunch's method, you will be getting the integer value for representing a character, not the actual value of the split numbers.

Why not try using a StringTokenizer for those numbers that are more than 1 digit in length?

[irmoathlete]

ok cool, i was on the right track i converted to the string object! =]

so using a string tokenizer i could just parse after i separated right?

[Crunch]

bcranger, on 22 July 2010 - 04:45 AM, said:

If you follow Crunch's method, you will be getting the integer value for representing a character, not the actual value of the split numbers.

Why not try using a StringTokenizer for those numbers that are more than 1 digit in length?

    int aInt = 9881;
    String aString = Integer.toString(aInt);
    int y = aString.charAt(0);    // <-- You will get the ascii value
    char x = aString.charAt(0);   // So do this
    System.out.println(aString.charAt(0));  // Or this to get the actual digit
    System.out.println(y);
    System.out.println(x);

[bcranger]

Crunch, on 21 July 2010 - 09:53 PM, said:

bcranger, on 22 July 2010 - 04:45 AM, said:

If you follow Crunch's method, you will be getting the integer value for representing a character, not the actual value of the split numbers.

Why not try using a StringTokenizer for those numbers that are more than 1 digit in length?

    int aInt = 9881;
    String aString = Integer.toString(aInt);
    int y = aString.charAt(0);    // <-- You will get the ascii value
    char x = aString.charAt(0);   // So do this
    System.out.println(aString.charAt(0));  // Or this to get the actual digit
    System.out.println(y);
    System.out.println(x);

Your method still does now allow him to add the two digits together, as adding any value returned using the charAt() method will esentially return to you the integer value of that character in ASCII. Adding "char x" to another, let's say "char y", will not give you, for instance "9", it will instead give you the sum of the two ASCII values. It DOES print out the values, but I believe he wanted to add them together.

[Crunch]

bcranger, on 22 July 2010 - 06:06 AM, said:

Crunch, on 21 July 2010 - 09:53 PM, said:

    int aInt = 9881;
    String aString = Integer.toString(aInt);
    int y = aString.charAt(0);    // <-- You will get the ascii value
    char x = aString.charAt(0);   // So do this
    System.out.println(aString.charAt(0));  // Or this to get the actual digit
    System.out.println(y);
    System.out.println(x);

Your method still does now allow him to add the two digits together, as adding any value returned using the charAt() method will esentially return to you the integer value of that character in ASCII. Adding "char x" to another, let's say "char y", will not give you, for instance "9", it will instead give you the sum of the two ASCII values. It DOES print out the values, but I believe he wanted to add them together.

  System.out.println(aString.charAt(0) -48);

You can ha

This post has been edited by Crunch: 22 July 2010 - 12:27 AM

[bcranger]

Yes, with that modification it would work

And he wanted the total sum, not the sum for every individual number

[cfoley]

I'm really not a fan of using text processing to work with numbers. Although the text code may look superficially simple, there is a LOT going on behind the scenes to make the compeuter deal with text. It's really not all that difficult with two simple maths operations:

Mod gives the remainder of a division so:

58173 % 10 = 3.

Integer division divides and rounds down

58173 / 10 = 5817

5817 % 10 = 7
5817 / 10 = 581

581 % 10 = 1
581 / 10 = 85

And so on.....

We can put this in a loop:

int number = 58173;
int sumDigits = 0;
while (number > 0) {
  int lastDigit = number % 10;
  sumDigits += lastDigit;
  number /= 10;
}

I'll leave it to you to put that in a method and have another loop to deal with several numbers.

[Dogstopper]

I think that method was far more robust than using text anyway...using characters can get really complicated...

@Crunch and bcranger, I'd use String if I were doing it your way, not characters. Something like this:

String stringNum = s.charAt(0) + ""; // Converts to String
int num = Integer.parseInt(stringNum);

That method is far easier than messing with chars

[macosxnerd101]

I agree with them about using chars, and maybe converting to ints based on the ASCII table. It's more memory efficient than creating a new String object each time you want to parse to an int, which is what happens because of String immutability. Albeit since you aren't using the new keyword to create your Strings, repeated digits will be pulled from the String pool.

[Dogstopper]

macosxnerd101, on 22 July 2010 - 07:50 AM, said:

I agree with them about using chars, and maybe converting to ints based on the ASCII table. It's more memory efficient than creating a new String object each time you want to parse to an int, which is what happens because of String immutability. Albeit since you aren't using the new keyword to create your Strings, repeated digits will be pulled from the String pool.

oooohhhh....right...Geez, how did I miss that one?!

[NeoTifa]

Idk if I'm missing something, but from skimming the thread, this is the solution I thought of. Idk if you can dereference a char, but Character has a .toString(), so I don't see why not. I don't have a compiler to test it for you.

String s = ""//list of numbers here
int sum = 0;
char c = '';

for(int i = 0; i < s.length(); i++) {
c = s.charAt(i);
if(c.isDigit()) {
sum += Integer.parseInt(c.toString()); //can't parse chars
} //end if
} //end for

//display sum

This code should loop through every character in the string, test if it's a digit, and if it is, turns it to a string to be parsed as an integer. Then it's added to the previous sum. Hope that helps ^__^

This post has been edited by NeoTifa: 22 July 2010 - 06:20 AM

[macosxnerd101]

@NeoTifa: That's more or less what Dogstopper just suggested. Except char doesn't have a toString(), so you'd need to use the Character wrapper class.

[Dogstopper]

NeoTifa, on 22 July 2010 - 08:19 AM, said:

Idk if I'm missing something, but from skimming the thread, this is the solution I thought of. Idk if you can dereference a char, but Character has a .toString(), so I don't see why not. I don't have a compiler to test it for you.

String s = ""//list of numbers here
int sum = 0;
char c = '';

for(int i = 0; i < s.length(); i++) {
c = s.charAt(i);
if(c.isDigit()) {
sum += Integer.parseInt(c.toString()); //can't parse chars
} //end if
} //end for

//display sum

This code should loop through every character in the string, test if it's a digit, and if it is, turns it to a string to be parsed as an integer. Then it's added to the previous sum. Hope that helps ^__^

You have so many syntax errors it's not even funny. You are trying to dereference a char. You need to use the static methods of the Character class.

Character.isDigit(c)

Character.toString(c)

And macosxnerd101 made a point that said constructing Strings is more time consuming than using chars. However, I think cfoley's solution was best.