What's the "java" way of converting chars (digits) to ints

Question

Given the following code:

    char x = '5';
    int a0 = x - '0'; // 0
    int a1 = Integer.parseInt(x + ""); // 1
    int a2 = Integer.parseInt(Character.toString(x)); // 2
    int a3 = Character.digit(x, 10); // 3
    int a4 = Character.getNumericValue(x); // 4
    System.out.printf("%d %d %d %d %d", a0, a1, a2, a3, a4);

(version 4 credited to: casablanca)

What do you consider to be the "best-way" to convert a char into an int ? ("best-way" ~= idiomatic way)

We are not converting the actual numerical value of the char, but the value of the representation.

Eg.:

convert('1') -> 1
convert('2') -> 2
....

I'm pretty sure anybody coming from those ancient C and Pascal things, like me, would vote for first option. — Nikita Rybak, Oct 16 '10 at 17:59
@Nikita: I do come from those "ancient C and Pascal things", but unless performance is absolutely critical, I would prefer a more expressive way than `x - '0'`. — casablanca, Oct 16 '10 at 18:11
Does this answer your question? [Java: parse int value from a char](https://stackoverflow.com/questions/4968323/java-parse-int-value-from-a-char) — user202729, Jan 27 '21 at 12:53

score 47 · Accepted Answer · answered Oct 16 '10 at 18:01

47

How about Character.getNumericValue?

answered Oct 16 '10 at 18:01

casablanca

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score 16 · Answer 2 · answered Oct 16 '10 at 18:00

16

I'd strongly prefer Character.digit.

answered Oct 16 '10 at 18:00

jacobm

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3

Please explain reasoning – James Wierzba Dec 04 '17 at 22:59

score 8 · Answer 3 · answered Oct 16 '10 at 18:00

8

The first method. It's the most lightweight and direct, and maps to what you might do in other (lower-level) languages. Of course, its error handling leaves something to be desired.

answered Oct 16 '10 at 18:00

Oliver Charlesworth

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score 4 · Answer 4 · answered Oct 19 '10 at 19:04

4

If speed is critical (rather than validation you can combine the result) e.g.

char d0 = '0';
char d1 = '4';
char d2 = '2';
int value = d0 * 100 + d1 * 10 + d2 - '0' * 111;

answered Oct 19 '10 at 19:04

Peter Lawrey

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I like the `'0' * 111` value :-) – dedek May 11 '15 at 12:51

score -1 · Answer 5 · answered Oct 15 '15 at 03:32

-1

Convert to Ascii then subtract 48.

(int) '7' would be 55 
((int) '7') - 48 = 7 
((int) '9') - 48 = 9

answered Oct 15 '15 at 03:32

ben

9

Kevin Ng · Answer 6 · 2019-02-04T23:04:58.767

-2

The best way to convert a character of a valid digit to an int value is below. If c is larger than 9 then c was not a digit. Nothing that I know of is faster than this. Any digits in ASCII code 0-9(48-57) ^ to '0'(48) will always yield 0-9. From 0 to 65535 only 48 to 57 yield 0 to 9 in their respective order.

int charValue = (charC ^ '0');

edited Feb 04 '19 at 23:04

answered Jan 25 '17 at 13:29

Kevin Ng

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What's the "java" way of converting chars (digits) to ints

6 Answers6