Getting ints array instead of characters array

Question

I am trying a code problem to convert double to string and then insert that to an array. I tried various methods but these don't give expected output.

public int[] makePi() {
  double PI = Math.PI;

  String sPI = String.valueOf(PI);
  int[] Arr = new int[3];

  for(int i =0; i<3; i++)
  {
    Arr[i] = sPI.charAt(i);
  }

  return Arr; 
}

Output should be an array with first three characters of PI as below :-

[ 3, 1, 4 ] while I am getting [51, 46, 49]

I will handle decimal character if needed.

Just a hint is needed.

Please don't provide full program that will be a spoiler. :-)

Character.isDigit and Character.getNumericValue will help you here. — Andy Turner
– Andy Turner, Commented Apr 25, 2016 at 7:33

Maroun · Accepted Answer · 2016-04-25 07:03:56Z

2

Look at the ASCII table. Do you see what are the corresponding chars for the integers you're getting? This should be a good hint for you.

Note that you're assigning the result to an int array, while you're running on characters.

answered Apr 25, 2016 at 7:03

Maroun

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7 Comments

RanchiRhino Over a year ago

I think this question(stupidity) doesn't meet the standard of SO.

Maroun Over a year ago

@RanchiRhino I hope there's no spoiler.

Maroun Over a year ago

@RanchiRhino It does. You described your problem and provided a reproducible code.

Tom Blodget Over a year ago

Not ASCII, Unicode/UTF-16 (numbers are hexadecimal and are in rows of 16.) From the question, 51 would be 0x33, which in UTF-16 is the single UTF-16 code unit needed for the Unicode codepoint U+0033. [I think it's better to learn that you know a little about Unicode than to think you need to know ASCII and that it is all you need to know.]

Maroun Over a year ago

@TomBlodget I'm actually very familiar with unicode/UTF-16. I even wrote a blogpost a while ago.

|

Raman Shrivastava · Accepted Answer · 2016-04-25 07:09:37Z

2

you're storing chars into an int array. hence theie respective ascii values will be stored in array (you're effectively converting char to int)

3 (char) -> 51 (ASCII Value)

. (char) -> 46 (ASCII Value)

1 (char) -> 49 (ASCII Value)

your array length is 3, so only first 3 chars are converted to ascii which is 3.1, not 3.14

But now if you want to store it into an char array (which i feel you're trying to do), all you need is -

char[] charArray = sPI.toCharArray();

Plus, I dont think you want to store in int array as though you can convert ascii values int their respective int value, but what about '.' which is not a valid int.

answered Apr 25, 2016 at 7:09

Raman Shrivastava

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Comments

Jędrzej Dudkiewicz · Accepted Answer · 2016-04-25 07:11:29Z

1

What you get in your array are values of characters (so something like 70 for '3', I neither remember nor want to remember exact values). You must convert value of character into the number itself. Hint: characters are numbered in the following way:

'0' - n

'1' - n + 1

'2' - n + 2

and so on.

edited Apr 25, 2016 at 7:11

answered Apr 25, 2016 at 7:03

Jędrzej Dudkiewicz

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Andy Turner · Accepted Answer · 2016-04-25 07:37:00Z

1

If you want to extract the numeric values of the digits, I would advise against doing explicit comparisons and arithmetic on the character values.

The Character class provides helper methods, which are less error-prone and more readable:

int outIndex = 0;
for (int i = 0; i < 3 /* && i < sPI.length() */; ++i) {
  char c = sPI.charAt(i);
  if (Character.isDigit(c)) {
    Arr[outIndex++] = Character.getNumericValue(c);
  }
}
/* assert outIndex == 3 */
return Arr;

I've commented out some code which I'd put in there for more robustness - it's not strictly necessary in this case, since we know that sPI has at least 3 digits in it. (Mind you, if we're going to hard-code that assumption, we may as well simply return new int[] { 3, 1, 4 };).

answered Apr 25, 2016 at 7:37

Andy Turner

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1 Comment

RanchiRhino Over a year ago

In that case (hard-coding), I should round the double to two decimal places.

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