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Could you help me understand how to get the same result by usint \n?

System.out.println("Balance: " + balance);
System.out.println();

I have tried something like 

System.out.println("Balance: " + balance +\n);

Not working. Don't know whether it is possible or not.

Kifsif
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6 Answers6

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System.out.println("Balance: " + balance +"\n");
Amilcar Andrade
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Use

System.out.println("Balance: " + balance +"\n");
rocketboy
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try

System.out.println("Balance: " + balance +"\n");
Zennichimaro
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You need to include "\n" in quotes. The "\" operator is to ignore the next operator, and n stands for new line. If you just put "n", it will be a string output of "n".

System.out.println("Balance: " + balance +"\n"); gives you:

Balance: *value* 

Next output will be here
wchargin
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user3871
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Put "\n" at the end of the string:

System.out.println("Balance: " + balance + "\n");

This works because \n is a special character that stands for newline. It's platform-independent; you won't need to use \r\n or anything like that. You can use \n anywhere a string is accepted.

However, in this case, I would actually favor the first approach, because it makes it clear what you're trying to do. A call to println() is much clearer than a \n subtly tacked on at the end there.

wchargin
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\n is the escape sequence for newline character and must be used as a constant of type char. So code will be:

System.out.println("Balance: " + balance + "\n");

or

System.out.println("Balance: " + balance +'\n');

There are other escape chars as \t (tab), \b etc Loeek here for full list and explanation

Luca Basso Ricci
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