Java - Char, Int Conversions

How can I convert a char to int in Java?

The ASCII table is arranged so that the value of the character '9' is nine greater than the value of '0'; the value of the character '8' is eight greater than the value of '0'; and so on.

So you can get the int value of a decimal digit char by subtracting '0'.

char x = '9';
int y = x - '0'; // gives the int value 9

Java implicit conversion between int and char

Just write the char conversion to ASCII code (below your statements)

int x = '0' + 1 - '5'
        48  + 1 - 53 = -4 

int y = '5' - 0 + '1'
         53 - 0 + 49 = 102

int y = '5' - '0' + '1'
         53 - 48  + 49 = 54

Notice it's consistent, each int remains int and each char converted to ASCII code

Java - char, int conversions

The first example (which compiles) is special because both operands of the addition are literals.

A few definitions to start with:

Converting an int to char is called a narrowing primitive conversion, because char is a smaller type than int.
'A' + 1 is a constant expression. A constant expression is (basically) an expression whose result is always the same and can be determined at compile-time. In particular, 'A' + 1 is a constant expression because the operands of + are both literals.

A narrowing conversion is allowed during the assignments of byte, short and char, if the right-hand side of the assignment is a constant expression:

In addition, if the expression [on the right-hand side] is a constant expression of type byte, short, char, or int:
A narrowing primitive conversion may be used if the variable is of type byte, short, or char, and the value of the constant expression is representable in the type of the variable.

c + 1 is not a constant expression, because c is a non-final variable, so a compile-time error occurs for the assignment. From looking at the code, we can determine that the result is always the same, but the compiler isn't allowed to do that in this case.

One interesting thing we can do is this:

final char a = 'a';
char b = a + 1;

In that case a + 1 is a constant expression, because a is a final variable which is initialized with a constant expression.

The caveat "if […] the value […] is representable in the type of the variable" means that the following would not compile:

char c = 'A' + 99999;

The value of 'A' + 99999 (which is 100064, or 0x186E0) is too big to fit in to a char, because char is an unsigned 16-bit integer.

As for the postfix ++ operator:

The type of the postfix increment expression is the type of the variable.
...
Before the addition, binary numeric promotion* is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion and/or subjected to boxing conversion to the type of the variable before it is stored.

(* Binary numeric promotion takes byte, short and char operands of operators such as + and converts them to int or some other larger type. Java doesn't do arithmetic on integral types smaller than int.)

In other words, the statement c++; is mostly equivalent to:

c = (char)(c + 1);

(The difference is that the result of the expression c++, if we assigned it to something, is the value of c before the increment.)

The other increments and decrements have very similar specifications.

Compound assignment operators such as += automatically perform narrowing conversion as well, so expressions such as c += 1 (or even c += 3.14) are also allowed.

Fastest way to convert numeric chars to int

The two versions are not equivalent:
- The Character.getNumericalValue(...) methods work for a variety of characters that represent digits or numbers, and it will return -1 or -2 in cases where the character doesn't represent a non-negative integer.
- The num - '0' approach only gives the correct answer for the codepoints that correspond to the ASCII characters '0' through '9'. For all other codepoints or codeunits, it gives a meaningless value.
The num - '0' version will be faster. This is clear from looking at the source code for getNumericalValue(...).
While the difference is significant in relative terms, it is very small in absolute terms.

I concur with the comments that say that this is most likely a premature optimization.

It is also an incorrect optimization in some contexts.

I use it a lot so was wondering if I was using the most efficient one.

This is definitely premature optimization :-)

The number of times you write a particular code sequence is unrelated to performance of the code sequence when is executed. A section of code is only worth optimizing if the time spent executing it makes a significant difference to your entire application.

Java char is also an int?

The Java Language Specification states

When a return statement with an Expression appears in a method
declaration, the Expression must be assignable (§5.2) to the declared
return type of the method, or a compile-time error occurs.

where the rules governing whether one value is assignable to another is defined as

Assignment contexts allow the use of one of the following:

a widening primitive conversion (§5.1.2)

and

19 specific conversions on primitive types are called the widening
primitive conversions:

char to int, long, float, or `double

and finally

A widening primitive conversion does not lose information about the
overall magnitude of a numeric value in the following cases, where the
numeric value is preserved exactly: [...]

A widening conversion of a char to an integral type T zero-extends the
representation of the char value to fill the wider format.

In short, a char value as the expression of a return statement is assignable to a return type of int through widening primitive conversion.

Java Implicit conversion char to int?

Array access expressions undergo implicit unary numeric promotion, which will widen an expression to int.

The index expression undergoes unary numeric promotion (§5.6.1).

The char datatype is widened to int via its Unicode value, e.g. 'A' -> 65.