byte a = 1;
byte b = 1;
byte c = a + b;

引发错误：可能会损失精度

byte subt = a_s - a_b;
                ^
  required: byte
  found:    int

这种行为是与jvm有关还是在java语言中定义的.

编辑：如果它是用java语言定义然后这样做,因为记住jvm？

意味着如果java支持byte数据类型,那么为什么对byte的操作会产生int

解决方法:

if java supports byte datatype then why operation on byte results int

因为这就是Java虚拟机的设计方式.没有指令集对字节类型执行操作.而是int类型的指令集用于对boolean,byte,char和short类型的操作.

从JVM Spec – Section 2.11.1开始：

A compiler encodes loads of literal values of types byte and short using Java Virtual Machine instructions that sign-extend those values to values of type int at compile-time or run-time. Loads of literal values of types boolean and char are encoded using instructions that zero-extend the literal to a value of type int at compile-time or run-time. [..]. Thus, most operations on values of actual types boolean, byte, char, and short are correctly performed by instructions operating on values of computational type int.

其背后的原因也在该部分中指定：

Given the Java Virtual Machine’s one-byte opcode size, encoding types into opcodes places pressure on the design of its instruction set. If each typed instruction supported all of the Java Virtual Machine’s run-time data types, there would be more instructions than could be represented in a byte. […] Separate instructions can be used to convert between unsupported and supported data types as necessary.

有关所有指令集可用于各种类型的详细信息,您可以查看该部分中的表.

还有一个表指定实际类型到JVM计算类型的映射：