45

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.

3
  • 4
    Use Horner-Scheme: ((((a * 10 + b) * 10 + c) * 10 + d) * 10 + e. Why do you lose b, when you add them up? Commented Apr 20, 2010 at 12:17
  • More on Horner-Scheme: stackoverflow.com/questions/1991380/… Commented Apr 23, 2010 at 6:34
  • I suppose you mean you "lose" a not b, iff a==0. Commented Jun 27, 2024 at 8:09

22 Answers 22

54

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.

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8 Comments

I'm not a fan of this - I think a + "" + b + "" + c + "" + d + "" + e would be clearer, myself - although more longwinded, of course. It's just too easy to look at "a + b" and think that will be adding the integers together, IMO.
@polygenelubricants: Thanks for pointing out that question - I hadn't spotted it before...
I think I would use a StringBuffer with chained appends.
@Andrei: What is the root of all evil again? Besides, one-line string concatenation will be compiled into a StringBuilder. It's only in more complex scenarios (usually involving loops) where concatenation should be avoided.
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37

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

Comments

36

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();
}

4 Comments

Thanks everyone for your answers, I'm always working with 5 digits, but in many cases they start with 0, so Michael's way's the wat way ;-)
This also works with 0s. For 5 digits, it always give the same result as Michael's; its only advantage (which you don't need) is that it works with variable number of digits.
Varargs + foreach combo is best.
this is cleaner and has better performance too... string concatenation is messy and results in a lot of unnecessary string generation. modern VMs may mitigate that... not sure, but i think this is much cleaner nonetheless.
17

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...)

2 Comments

this is the best solution so far
Or, if you are printing to the console, System.out.printf("%d%d%d%d%d", a, b, c, d, e);
13

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.

1 Comment

Consider that considered, added it to the note.
7

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);

1 Comment

the question was: "How do i concatenate these values so that i end up with a String..."!
5
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.

2 Comments

StringBuilder in this case is an overkill; plain old + is fine and much more readable.
Readable for who? I prefer the StringBuffer/StringBuilder in this case, and agreeing with Jon Skeets comment to Michaels answer.
5

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

2 Comments

what if my first value is equal to 0?
It would still work. It will give you the proper value. Do you need it to be exactly 5 chars long?
4

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);

Comments

4

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.

Comments

3

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());

1 Comment

Or : Stream.of(a, b, c, d, e).map(Object::toString).collect(Collectors.joining()); Or: IntStream.of(a, b, c, d, e).mapToObj(Integer::toString).collect(Collectors.joining());
2
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.

Comments

1

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());

4 Comments

He seems to want a String as the end result, so the parsing is unnecessary (and if that were actually wanted, the correct solution would be to just add up the numbers without any String tomfoolery)
Good point on him wanting a String, but if the numbers were just added, the result would be 6, not 10221...
@polygenelubricants: Totally agree with you - forget this answer, Michael or Jon's answer are the best solutions going on the information given :)
@polygenelubricants: ... or yours ;-)
1

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());

Comments

1

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));

Comments

1

You can Use

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

1 Comment

You can just "" + a + b + c
1

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);

Comments

0

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. :)

Comments

0

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) ;
}

Comments

0
//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);  
    } 
}

Comments

-1

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

Comments

-4
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);  
    }
}

Comments

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