Question

I’m having trouble figuring out the inner workings of switches in Java
I’m told that for all primitives, the value is promoted to a Integer.

However, in the following example, I’m testing on a byte variable, and any case larger than 127 will not compile:

byte k = 5;
switch(k){
  case 128:    //fails to compile, possible loss of precision  

I realize this is an error and have no issue with that. My question is:
How does the JVM track that it’s switching on a byte if it takes the value of “k” and promotes it to an integer before testing each case?

Answer 1

The problem is it’s not promoting the value of k, it’s trying to take the case statement (128 – signed integer) and assign it to a byte. As 128 is larger than 1 byte (7 bits + sign bit) then the compilation fails.

For example

byte k = 128; 

would also fail to compile.
See the Java Language Specification