Java: Subtract '0' from char to get an int... why does this work?

Question

This works fine:

int foo = bar.charAt(1) - '0';

Yet this doesn't - because bar.charAt(x) returns a char:

int foo = bar.charAt(1);

It seems that subtracting '0' from the char is casting it to an integer.

Why, or how, does subtracting the string '0' (or is it a char?) convert another char in to an integer?

• Mark Peters: Don't make the mistake of thinking that '0' == 0. In reality, '0' == 48.
• ledneb: click Ah, now it makes perfect sense! If only you'd put that as an answer and pointed me to cs.utk.edu/~pham/ascii_table.jpg
• tonix: this implicit cast doesn't return the decimal representation of the char, as @MarkPeters pointed, so if for example bar.charAt(1) returns 'A', which dec representation is 65, int foo will be 17, 'A' - '0' = 65 - 48 = 17 and not 65.
• Aaron Liu: This trick is being used in a lot of questions in Leetcode, but I don't see anyone asked about how this trick works for Leetcode questions.

Answer 1

That's a clever trick. char's are actually of the same type / length as shorts. Now when you have a char that represents a ASCII/unicode digit (like '1'), and you subtract the smallest possible ASCII/unicode digit from it (e.g. '0'), then you'll be left with the digit's corresponding value (hence, 1)

Because char is the same as short (although, an unsigned short), you can safely cast it to an int. And the casting is always done automatically if arithmetics are involved

• OscarRyz: shorts can have negative values, char doesn't. Also is not ASCII but unicode ( small but important difference ) : try int y = '\uffff' - '\ufffe';
• Lukas Eder: You're right. For digits, ASCII and Unicode codes happen to coincide, that's why I thought of ASCII first
• ledneb: That's great. I did need Mark Peter's comment to make me realise that by 'corresponding value' you mean I'm actually subtracting 48 from 49-57. Also, the error I was getting was really an array out of bounds exception so perhaps my question's misleading in saying the first example "doesn't work". Thanks! :)
• ColinD: To be clear, a char isn't considered a subtype of short in Java... your answer kind of makes it sound like it is. char and short are both direct subtypes of int. Also, "casting" is done automatically between any type and any type that is considered a supertype of it regardless of whether arithmetic is involved.

Answer 2

This is an old ASCII trick which will work for any encoding that lines the digits '0' through '9' sequentially starting at '0'. In Ascii, '0' is a character with value 0x30 and '9' is 0x39. Basically, if you have a character that is a digit, subtracting '0' "converts" it to it's digit value.

I have to disagree with @Lukas Eder and suggest that it is a terrible trick; because the intent of this action aproaches 0% obvious from code. If you are using Java and have a String that contains digits and you want to convert said String to an int I suggest that you use Integer.parseInt(yourString);.

This technique has the benifit of being obvious to the future maintenance programmer.

• Lukas Eder: it's a clever trick. and clever is always a bad thing. it's the opposite of elegant :-)

Answer 3

'0' is a char too. It turns out, the characters in Java have a unicode (UTF-16) value. When you use the - operator with characters Java performs the operation with the integer values.

For instance, int x = 'A' - '0';// x = 17

• Mark Peters: It doesn't complain at all. You can do int y = s.charAt(1) just fine, without compile or runtime error (assuming s has two+ characters). I think the OP was confused as to why it didn't return the same result.
• Mark Peters: @Bart: UTF-16 is an encoding format for unicode, but yes you're right.
• Bart van Heukelom: @Mark: I know, but for characters above unicode point 65k, (at least) two UTF-16 char's are used
• Tom Blodget: You chose a great example to illustrate to limited usefulness of character arithmetic.
• Zhou Haibo: With above example , I get jshell> int x = '0' - 'A' x ==> -17.

Answer 4

chars are converted to int implicitly:

 public static void main(String [] args) throws Exception {
 String bar = "abc";
 int foo = bar.charAt(1) - '0';
 int foob = bar.charAt(1);
 System.err.println("foo = " + foo + " foob = " + foob);
 }

output: foo = 50 foob = 98.

Maybe you put two int foo ... and this is because it didn't work?

Answer 5

The following code works perfectly fine!

int foo = bar.charAt(1);

Similar to reference types, any Java primitive can be assigned without casting to another primitive of a type it is considered a subtype of. The subtyping rules for primitives are given by JLS section 4.10.1. char is considered a subtype of int, so any char may be assigned to an int.

• Alex Fedulov: Are you sure? String s = "9"; int n = s.charAt(0); System.out.println(n); returns 57

Answer 6

Your code may compile without error & run without throwing an exception, but converting between char's & int's is bad practice. First, it makes the code confusing, leading to maintenance headaches down the road. Second, clever "tricks" can prevent compilers from optimizing the byte code. One of the best ways to get fast code is to write dumb code (i.e., not clever code).

Answer 7

Because in Java if you do not specify a type indicator with a number then it assumes Integer. What I mean by that, if you want to set a Long value, it would need to be 0L.

Performing a numerical operation on two different numerics, the result takes the larger. Hence, a char (which is a numerical value) minus an integer, results in an integer.

You will find that this works also

long val = bar.charAt(1) - 0L;

• khachik: @Lukas it won't, see my answer.

Answer 8

Your second snipped should work fine though:

int foo = bar.charAt(1);

A char, just like a short or byte, will always be silently cast to int if needed.

This will compile just fine: int i = 'c';

• Dmitri: Not saying it is, just that you don't need to explicitly cast char to int, so it's not clear why searbe would need the subtraction.
• Vasile Surdu: nah, if i have a "char c=bar.charAt(1); //c=1;" then "int foo=c;" foo will be 49 not 1. . So subtracting "int foo=c-'0'; returns 1. :) because 49-48=1

Answer 9

I will echo what @Mark Peters has said above in case people overlook his comment.

As I quote: " Don't make the mistake of thinking that '0' == 0. In reality, '0' == 48 "

Answer 10

'5' has the int value of 53

if we write '5'-'0' it evaluates to 53-48, or the int 5

if we write char c = 'B'+32; then c stores 'b'

'b' = 98.

Please refer ASCII chart https://en.cppreference.com/w/cpp/language/ascii