带有“表达式”的Java类型提升

Question

Can the variable below (called b) be called an expression, if it is the only thing located to right of the equals sign?

// This code fragment will not compile.
// c is a char, and b is a byte.

c = b;

The reason why I ask this question is because of the topic of type promotion within expressions. I understand that Java promotes all bytes to ints. Is that the only reason why this code fragment does not compile? (Please note that I know about casts; that's not the point of this thread. Thanks a lot.)

Edit: Thanks a lot Jon and Peter. Looking at this topic using a second example:

byte b = 1;
short s = 2;

s = b;  // OK
s = b*2;  // Not OK (compilation error)

Is it true that the following is happening?

(Line 3) Java converts the byte to short. (Line 4) Java converts the expression b*2 to an int.

If this is correct, then it would seem that =b; and =b*2; are "expressions" that Java handles differently. So, the =b; "expression" is not converted to an int, but widened to a short. But the =b*2; expression is converted to an int, and not to a short, even though the target variable called s is a short.

Edit 2: Also -

short s1, s2 = 2, s3 = 2;
s1 = s2*s3;  // Not OK (compilation error)

Even though all three variables are shorts, the s2*s3; expression is promoted to an int, and thus causes a compilation error.

Answer 1

Try this

byte b = -1;
short s = b; // is ok as a byte can be promoted to an short.
int i = b; // is ok as a byte can be promoted to an int.
float f = b; // is ok as a byte can be promoted to an float, long or double.

char c = b; // won't compile

but

final byte b = 1;
char c = b; // compiles fine as the compiler can inline the value.

---

In this case

short s = b*2;  // Not OK (compilation error)

b * 2 is an int as 2 is an int value. If b were final you would be able to do this was the compile can inline the value.

Answer 2

Can the variable below (called b) be called an expression, if it is the only thing located to right of the equals sign?

Yes, absolutely.

I understand that Java promotes all bytes to ints.

Well, in some cases. Not in all cases.

Fundamentally the code doesn't compile because there's no implicit conversion from byte to char . From section 5.1.2 of the JLS (widening primitive conversions):

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

• byte to short , int , long , float , or double
• ...

Note the lack of char in the list of target types for conversions from byte .