How to get toString for each element of an ArrayList?

Question

I have this class has constructor taking a String and a Arraylist as argument;

class Foo{
    String ide;
    ArrayList list;

    public TypeOperator(String ide, ArrayList list){
        this.ide = ide;
        this.list = list;
    }
}

so when I initialise the class like below:

ArrayList<SomeType> list = new ArrayList<Some Type>;
list.add(SomeType1);
list.add(SomeType2);
Foo one = new Foo("->", list);

is there a way to print the format like below:

SomeType 2 -> SomeType 1

Let's say I have a SomeType class. Do I have something to do in the SomeType class?

It's not clear what you want to do here. Are you looking to iterate over all the elements in the array and print out their toString() value? — Paul MacGuiheen, Jul 10 '16 at 01:07
Do you want to write Foo.toString() by iterating over values or you want to override toString of ArrayList? Take a look at: http://stackoverflow.com/questions/18129505/how-can-i-override-the-tostring-method-of-an-arraylist-in-java — Ali Rokni, Jul 10 '16 at 01:15

score 1 · Answer 1 · answered Jul 10 '16 at 01:15

First of all, it's highly recommended to stay away from raw types. List is a generic type, so you should use it as a generic type.

Then, what you want to do is a mere join operation. If you are using Java 8, you can do

list.stream().collect(Collectors.joining(" -> "))

In plain old Java, you can do

StringBuilder sb = new StringBuilder();
Iterator<T> it = list.iterator();

if (it.hasNext()) sb.append(it.next());
while (it.hasNext()) {
    sb.append(" -> ").append(it.next());
}
String result = sb.toString();

minioim · Answer 2 · 2016-07-10T11:05:31.217

1

You need to implement the toString() method from Foo which will be something like:

public String toString(){
    String result  = "" ;
    for(int  i = list.size()-1;i>0;i--){
        result = result+ list.get(i).toString() + " " + ide + " ";
    }
    result = result + list.get(0).toString();
    return result;
}

or, with better performances, as Dici pointed out :

    public String toString(){


    StringBuilder result  = new StringBuilder() ;
    if (!list.isEmpty()){
        for(int  i = list.size()-1;i>0;i--){
            result.append(list.get(i).toString()).append(" ").append(ide).append(" ");
        }
        result.append(list.get(0).toString());
    }
    return result.toString();
}

Then following the way you want to display SomeType you will probably have to implement its own toString method.

edited Jul 10 '16 at 11:05

answered Jul 10 '16 at 01:16

minioim

556
3
18

String concatenation is slow when done repeatedly to aggregate a result. `StringBuilder` is much faster. Also, your code will fail for an empty list – Dici Jul 10 '16 at 01:19
It's been a while since java compiler handle conversion from + operator to StringBuilder if it improves performance. http://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.18.1. Yes this code will fail in case of empty list. That's why i said "will be something like" ;) – minioim Jul 10 '16 at 01:30
You should not rely on the compiler to write good code in your place. Plus, you're wrong. The compiler will optimize `a + b + c` in a single expression but will not (or may not, depending on the implementaiton) optimize for successive statements with multiple reassignments. I'll create a snippet to prove it to you – Dici Jul 10 '16 at 03:39
my bad, i tried it by myself. I thought it would work even with no single expression. anyway I agree about the fact that rely on the compiler is not a good idea. this was only a simple example, easier to understand than StringBuilder. – minioim Jul 10 '16 at 11:00
No problem. As a general rule, I would not favour a "simple answer" over the "right answer". Part of Stack Overflow's role is to give people access to the best practices, not just the straightest solution to their problem – Dici Jul 10 '16 at 14:58
1

Right. Will remember it for next time (i'm juste starting to answer on SO) – minioim Jul 10 '16 at 15:00
That's ok, you are willing to take feedback and incorporate it in your answer, which is great. Anyway, good stack-overflowing :p – Dici Jul 10 '16 at 15:02

score 0 · Answer 3 · answered Jul 10 '16 at 01:12

You need create a method toElementString in Foo class.

 public String toElementString(){
        if(list == null || list.size() == 0) return "";
        StringBuffer buffer = new StringBuffer();
        int size = list.size();
        for(int i = 0; i < size;i++){
            buffer.append(list.get(i));
            if(i != (size - 1)){
                buffer.append(" ");
                buffer.append(ide);
                buffer.append(" ");
            }
        }
        System.out.println(buffer.toString());
        return buffer.toString();
    }

The empty list is not a special case since it will never enter the loop. Also, `StringBuider` is preferred over `StringBuffer` for non-concurrent usages. Finally, you will end up with the separator at the end of the string, which is undesired (something like `item1 sep item2 sep item3 sep`) — Dici, Jul 10 '16 at 01:17

How to get toString for each element of an ArrayList?

3 Answers3