How to concatenate int values in java?

Question

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.

Answer 1

//Here is the simplest way 
public class ConcatInteger{
    public static void main(String[] args) {
      
      int [] list1={1,2,3};
      int [] list2={1,9,6};
      
      String stNum1="";
      String stNum2="";
   
      for(int i=0 ; i<3 ;i++){
        stNum1=stNum1+Integer.toString(list2[i]);   //Concat done with string  
      }
      
      for(int i=0 ; i<3 ;i++){
        stNum2=stNum2+Integer.toString(list1[i]);
      }
      
      int sum= Integer.parseInt(stNum1)+Integer.parseInt(stNum2); // Converting string to int
      
         System.out.println(sum);  
    } 
}

Answer 2

NOTE: when you try to use + operator on (string + int) it converts int into strings and concatnates them ! so you need to convert only one int to string

public class scratch {

public static void main(String[] args){

    int a=1;
    int b=0;
    int c=2;
    int d=2;
    int e=1;
    System.out.println( String.valueOf(a)+b+c+d+e) ;
}

Answer 3

int number =0;
int[] tab = {your numbers}.

for(int i=0; i<tab.length; i++){
    number*=10;
    number+=tab[i];
}

And you have your concatenated number.

Answer 4

Using Java 8 and higher, you can use the StringJoiner, a very clean and more flexible way (especially if you have a list as input instead of known set of variables a-e):

int a = 1;
int b = 0;
int c = 2;
int d = 2;
int e = 1;
List<Integer> values = Arrays.asList(a, b, c, d, e);
String result = values.stream().map(i -> i.toString()).collect(Collectors.joining());
System.out.println(result);

If you need a separator use:

String result = values.stream().map(i -> i.toString()).collect(Collectors.joining(","));

To get the following result:

1,0,2,2,1

Edit: as LuCio commented, the following code is shorter:

Stream.of(a, b, c, d, e).map(Object::toString).collect(Collectors.joining());

Answer 5

public class joining {

    public static void main(String[] args) {
        int a=1; 
        int b=0;
        int c=2;
        int d=2;
        int e=1;

        String j = Long.toString(a);
        String k = Long.toString(b);
        String l = Long.toString(c);
        String m = Long.toString(d);
        String n = Long.toString(e);

       /* String s1=Long.toString(a);    // converting long to String
        String s2=Long.toString(b);
        String s3=s2+s1;
        long c=Long.valueOf(s3).longValue();    // converting String to long
        */

        System.out.println(j+k+l+m+n);  
    }
}

Answer 6

Couldn't you just make the numbers strings, concatenate them, and convert the strings to an integer value?

Answer 7

Assuming you start with variables:

int i=12;
int j=12;

This will give output 1212:

System.out.print(i+""+j); 

And this will give output 24:

System.out.print(i+j);

Answer 8

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

10000
    0
  200
   20
    1
=====
10221

Answer 9

This worked for me.

int i = 14;
int j = 26;
int k = Integer.valueOf(String.valueOf(i) + String.valueOf(j));
System.out.println(k);

It turned out as 1426

Answer 10

How about not using strings at all...

This should work for any number of digits...

int[] nums = {1, 0, 2, 2, 1};

int retval = 0;

for (int digit : nums)
{
    retval *= 10;
    retval += digit;
}

System.out.println("Return value is: " + retval);

Answer 11

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 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

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 14

You can Use

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

Answer 15

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 16

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 17

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 18

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 19

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 20

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 21

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 22

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.