How I can use an external variable in a lambda expression

Question

I'm trying to use a lambda expression in a forEach loop to concatenate to a variable "i" a string that is evaluated for each object.

here is the code

private String getStringActiveRooms(@NotNull ArrayList<Chat_room> c){
    String i;
    c.forEach( (chat_room) -> i = i.concat(chat_room.getName() + "[" + chat_room.activeUsers() + "/" + chat_room.maxUsers() + "]" + ", "));

    return i;

}

It throws me an error because external variables must be final to be used in a lambda expression.
But if I make final the variable I can't change it and the code didn't work.

Does anyone have a solution?

Does this answer your question? [Variable used in lambda expression should be final or effectively final](https://stackoverflow.com/questions/34865383/variable-used-in-lambda-expression-should-be-final-or-effectively-final) — Stubborn, Jan 13 '20 at 20:53
Why not map over each object in `c` and create a new arraylist and then join all the values together? [Related SO question](https://stackoverflow.com/questions/31523935/selecting-objects-from-an-arraylist-based-on-value-of-certain-fields) — Sunny Patel, Jan 13 '20 at 20:53
Do [these questions answer your question](https://www.google.com/search?q=java+lambda+use+outside+variable+site:stackoverflow.com)? What searching have you done prior to asking this question? — Hovercraft Full Of Eels, Jan 13 '20 at 20:54
first I was using a for loop but now I know lambda and I tried a new form — ThaDome23, Jan 13 '20 at 20:56
I found solution making it the first elemento of a final array — ThaDome23, Jan 13 '20 at 21:03
They are fine for your purpose. The heavy operations here are the string manipulations. — Thorbjørn Ravn Andersen, Jan 13 '20 at 21:10

Renato · Accepted Answer · 2020-01-13T21:04:33.180

You can use StringBuffer

private String getStringActiveRooms(@NotNull ArrayList<Chat_room> c){
    final StringBuffer i = new StringBuffer();
    c.forEach( (chat_room) -> i.append(chat_room.getName() + "[" + chat_room.activeUsers() + "/" + chat_room.maxUsers() + "]" + ", "));

    return i.toString();

}

In addition you have to know that using StringBuffer is better from a performance point of view: String are immutable this means that if you have a String i every iteration i + "something" creates a new String object in the heap space allocating memory for a temporary variable

How I can use an external variable in a lambda expression

1 Answers1