Java reserved name compilation

Question

I have seen this cod, why it works please?

public void nеw() {
    System.out.println("!?");
}

It is compile in java, I think new is a reserved word

One of these questions pops up every few weeks. I think the people who post them think they're asking a clever riddle, haha. — asteri, Jan 13 '15 at 22:06

score 10 · Accepted Answer · edited May 23 '17 at 11:57

10

It is a cyrillic е, not a latin e. Therefore, it looks like new, but is actually a different method name.

Like NPE said, I would recommend reading is it a good idea to use unicode symbols as java identifiers?

edited May 23 '17 at 11:57

Community

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answered Jan 13 '15 at 21:59

416E64726577

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score 7 · Answer 2 · edited May 23 '17 at 11:50

7

Look at the hex dump:

$ xxd -g 1 code.java
0000000: 70 75 62 6c 69 63 20 76 6f 69 64 20 6e d0 b5 77  public void n..w
                                                ↑↑ ↑↑                  ↑↑
0000010: 28 29 20 7b 0a 20 20 20 20 53 79 73 74 65 6d 2e  () {.    System.
0000020: 6f 75 74 2e 70 72 69 6e 74 6c 6e 28 22 21 3f 22  out.println("!?"
0000030: 29 3b 0a 7d 0a 0a                                );.}..

The "e" is not the Latin "e". It is a CYRILLIC SMALL LETTER IE (U+0435 encoded in UTF-8). Therefore the nеw in the code is not the new keyword, but is an identifier that happens to look like the new keyword.

Take a look at Is it a good idea to use unicode symbols as Java identifiers?

edited May 23 '17 at 11:50

Community

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answered Jan 13 '15 at 21:58

NPE

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Thanks for showing how you determined this. – Josh Jan 13 '15 at 22:01
That proves what they say about programming languages: "Once they go Latin, ..." – hfontanez Jan 13 '15 at 22:02

Java reserved name compilation

2 Answers2