How to concatenate int values in java?

Question

hi,

I have the following values:

int a=1; 
int b=0;
int c=2;
int d=2;
int e=1;

How do i concatenate these values so that i end up with a String that is 10221; please note that multiplying a by 10000, b by 1000.....and e by 1 will not working since b=0 and therefore i will lose it when i add the values up.

Thnks you in advance.

Answer 1

Actually,

int result = a * 10000 + b * 1000 + c * 100 + d * 10 + e;
String s = Integer.toString(result);

will work.

Note: this will only work when a is greater than 0 and all of b, c, d and e are in [0, 9]. For example, if b is 15, Michael's method will get you the result you probably want.

Answer 2

If you multiply b by 1000, you will not loose any of the values. See below for the math.

10000
    0
  200
   20
    1
=====
10221

Answer 3

StringBuffer sb = new StringBuffer();
sb.append(a).append(b).append(c)...

Keeping the values as an int is preferred thou, as the other answers show you.

Answer 4

I would suggest converting them to Strings.

StringBuilder concatenated = new StringBuilder();
concatenated.append(a);
concatenated.append(b);
/// etc...
concatenated.append(e);

Then converting back to an Integer:

Integer.valueOf(concatenated.toString());

Answer 5

Use StringBuilder

StringBuilder sb = new StringBuilder(String.valueOf(a));
sb.append(String.valueOf(b));
sb.append(String.valueOf(c));
sb.append(String.valueOf(d));
sb.append(String.valueOf(e));
System.out.print(sb.toString());

Answer 6

Others have pointed out that multiplying b by 1000 shouldn't cause a problem - but if a were zero, you'd end up losing it. (You'd get a 4 digit string instead of 5.)

Here's an alternative (general purpose) approach - which assumes that all the values are in the range 0-9. (You should quite possibly put in some code to throw an exception if that turns out not to be true, but I've left it out here for simplicity.)

public static String concatenateDigits(int... digits)
{
    char[] chars = new char[digits.length];
    for (int i = 0; i < digits.length; i++)
    {
        chars[i] = (char)(digits[i] + '0');
    }
    return new String(chars);
}

In this case you'd call it with:

String result = concatenateDigits(a, b, c, d, e);

Answer 7

The easiest (but somewhat dirty) way:

String result = "" + a + b + c + d + e

Edit: I don't recommend this and agree with Jon's comment. Adding those extra empty strings is probably the best compromise between shortness and clarity.

Answer 8

Michael Borgwardt's solution is the best for 5 digits, but if you have variable number of digits, you can use something like this:

public static String concatenateDigits(int... digits) {
   StringBuilder sb = new StringBuilder(digits.length);
   for (int digit : digits) {
     sb.append(digit);
   }
   return sb.toString();
}

Answer 9

People were fretting over what happens when a == 0. Easy fix for that...have a digit before it. :)

int sum = 100000 + a*10000 + b*1000 + c*100 + d*10 + e;
System.out.println(String.valueOf(sum).substring(1));

Biggest drawback: it creates two strings. If that's a big deal, String.format could help.

int sum = a*10000 + b*1000 + c*100 + d*10 + e;
System.out.println(String.format("%05d", sum));

Answer 10

You can Use

String x = a+"" +b +""+ c+""+d+""+ e;
int result = Integer.parseInt(x);

Answer 11

For fun... how NOT to do it ;-)

String s = Arrays.asList(a,b,c,d,e).toString().replaceAll("[\\[\\], ]", "");

Not that anyone would really think of doing it this way in this case - but this illustrates why it's important to give access to certain object members, otherwise API users end up parsing the string representation of your object, and then you're stuck not being able to modify it, or risk breaking their code if you do.

Answer 12

just to not forget the format method

String s = String.format("%s%s%s%s%s", a, b, c, d, e);

(%1.1s%1.1s%1.1s%1.1s%1.1s if you only want the first digit of each number...)

Answer 13

Best solutions are already discussed. For the heck of it, you could do this as well: Given that you are always dealing with 5 digits,

(new java.util.Formatter().format("%d%d%d%d%d", a,b,c,d,e)).toString()

I am not claiming this is the best way; just adding an alternate way to look at similar situations. :)

Answer 14

you can use bit shift operation as well here no?