Puzzle of compound assignment operator

Let's look at an example:

// Add a declaration  while I = 0) {      i>>>=1;       }

Add a line of code at the Add Declaration to make the above loop a dead loop

Long i =-1;

or int =-1;

Oh, no!!!

Try short i =-1;

What's up?

Dead cycle!!!!

Since there's been a strange phenomenon (at least it's weird for me)

Why is it?

Lookup data A specification for complex assignment operations

In a compound assignment operation, a type that is less than int is promoted by an int type (that is, widening), then computed, and then narrowed to the specified operation.

That is, when the type of I is short, it is converted to type int (0xFFFFFFFF), then to the right (0x7FFFFFFF), and then to Short (0xFFFFFFFF).

Then watch the following code:

 Short x = 1,short y = 2, Short z = x + y;

Run discovery

Thirtyfirst.java:7: Error: Possible loss of        precision Short z = x + y                    ;  ^  Short   found:    int1 errors

In other words, there is an implicit extension to int, which then needs to be forced to a short type

Summarize:

<1> when arithmetic operations are performed, if the values on both sides of the operator are lower than int (that is, the byte or short type is automatically converted to the int type first)

<2>, at the time of the operation, will be automatically converted to the high level of the one to do the operation, so the result is also a high level of that, if you want to lower the need to force type conversion

Puzzle of compound assignment operator