java - java中的隐式缩小规则

Question

byte b = 0xFFFFFFFF; //OK, because integer -1 sits between -128 and 127, FINE!!
char ch = 0xFFFFFFFF; //Not OK, because integer -1 does not sit between 0 and 65535, FINE!!
byte b = 0L; //Compiler says Not OK? But long integer 0 sits between -128 and 127?

我不相信 java 编译器在上面代码的第三行中应用的缩小规则。

请帮助我理解这个缩小规则背后的逻辑。

Answer 1

文字 0L 上的 L 后缀使此文字成为 long 类型(64 位有符号整数)。

根据 Java 语言的规则，没有从 long 到 byte 的隐式缩小。

参见 Java 语言规范 section 5.2 Assignment Contexts :

In addition, if the expression is a constant expression (§15.28) of type byte, short, char, or int:

• A narrowing primitive conversion may be used if the type of the variable is byte, short, or char, and the value of the constant expression is representable in the type of the variable.

注意常量表达式的类型不包括long。