up to first N characters

Question

How do I get up to the first n characters of a string in java without doing a size check first(inline is acceptable) or risking an IndexOutOfBoundsException?

Answer 1

Use the substring method, as follows:

int n = 8;
String s = "Hello, World!";
System.out.println(s.substring(0,n);

If n is greater than the length of the string, this will throw an exception, as one commenter has pointed out. one simple solution is to wrap all this in the condition if(s.length()<n) in your else clause, you can choose whether you just want to print/return the whole String or handle it another way.

Answer 2

Here's a neat solution:

String upToNCharacters = s.substring(0, Math.min(s.length(), n));

Answer 3

There's a class of question on SO that sometimes make less than perfect sense, this one is perilously close :-) Perhaps you could explain your aversion to using one of the two methods you ruled out.

If it's just because you don't want to pepper your code with if statements or exception catching code, one solution is to use a helper function that will take care of it for you, something like:

static String substring_safe (String s, int start, int len) { ... }

which will check lengths beforehand and act accordingly (either return smaller string or pad with spaces).

Then you don't have to worry about it in your code at all, just call:

String s2 = substring_safe (s, 10, 7);

instead of:

String s2 = s.substring (10,7);

This would work in the case that you seem to be worried about (based on your comments to other answers), not breaking the flow of the code when doing lots of string building stuff.