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.

Frizzly

Use Horner-Scheme: ((((a * 10 + b) * 10 + c) * 10 + d) * 10 + e. Why do you lose b, when you add them up? – Nickolai

More on Horner-Scheme: #1991880 – Bobstay

I suppose you mean you "lose" a not b, iff a==0. – Parch

Answer 1

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.

Irony

Dirty? What? #2506974 – Bobstay

@polygenelubricants: well, I happen to disagree with the majority opinion on that point. – Irony

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. – Conrad

@polygenelubricants: Thanks for pointing out that question - I hadn't spotted it before... – Conrad

I think I would use a StringBuffer with chained appends. – Tumer

Isn't direct string concatenation a little inefficient compared to StringBuffer/Builder methods ? – Zymo

@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. – Irony

What about "" + (a) + (b) + (c) + (d) + (e)? Just as concise, and it draws just enough attention to itself that you go "Ah, I see what's going on here...". – Bobstay

Answer 2

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

Natashianatassia

Answer 3

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();
}

Bobstay

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 ;-) – Frizzly

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. – Bobstay

Varargs + foreach combo is best. – Bobstay

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. – Amatruda

Answer 4

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...)

Headmost

this is the best solution so far – Tetragrammaton

Or, if you are printing to the console, System.out.printf("%d%d%d%d%d", a, b, c, d, e); – Adeno

Answer 5

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.

Sisile

Consider that considered, added it to the note. – Sisile

Answer 6

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);

Ibex

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

This helped me with a palindrome problem, Thanks! – Juta

Answer 7

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

Blastopore

what if my first value is equal to 0? – Frizzly

It would still work. It will give you the proper value. Do you need it to be exactly 5 chars long? – Blastopore

Answer 8

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.

Padishah

StringBuilder in this case is an overkill; plain old + is fine and much more readable. – Bobstay

Readable for who? I prefer the StringBuffer/StringBuilder in this case, and agreeing with Jon Skeets comment to Michaels answer. – Mailemailed

Answer 9

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);

Conrad

Answer 10

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.

Asparagus

Answer 11

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());

Sapowith

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()); – Perspicacity

Answer 12

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.

Showoff

Answer 13

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());

Sparklesparkler

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) – Irony

Good point on him wanting a String, but if the numbers were just added, the result would be 6, not 10221... – Sparklesparkler

@polygenelubricants: Totally agree with you - forget this answer, Michael or Jon's answer are the best solutions going on the information given :) – Sparklesparkler

@polygenelubricants: ... or yours ;-) – Sparklesparkler

Answer 14

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());

Shiner

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));

Mousey

Answer 16

You can Use

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

Abeyance

You can just "" + a + b + c – Lashay

Answer 17

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);

Isosceles

Answer 18

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. :)

Gilges

Answer 19

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) ;
}

Dewberry

Answer 20

//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);  
    } 
}

Morsel

Answer 21

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

Autocephalous

Answer 22

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);  
    }
}

Incoherent