在java中移位时的不同行为

Question

I'm working with bit shifting in Java and have the following piece of code that works as expected:

final byte value = 1;
final int shift = 1;
byte result = value << shift;

This produces the value 2 as expected. If however I attempt to extract this into a method like so:

private void shiftAndCheck(final byte value, final int shift) {
  byte result = value << shift;
}

This results in a compilation error:

java: incompatible types: possible lossy conversion from int to byte

The question is what is it about the method that causes this to fail?

Answer 1

Since the value and shift are compile-time constants in this snippet:

final byte value = 1;
final int shift = 1;
byte result = value << shift;

then the compiler inlines their values (replaces all the occurrences of value and shift with their actual values of 1 ) and can verify before Runtime that the result of value << shift won't cause any loss of precision.

---

Meanwhile, for the second snippet:

private void shiftAndCheck(final byte value, final int shift) {
  byte result = value << shift;
}

the compiler has no evidence that shift will represent such value, which wouldn't cause loss of precision and that's why raises a compilation error.

If compilation was possible in this case, then you'd have been able to do shiftAndCheck(1, 32); which would result in overflowing the byte type.

Answer 2

默认情况下，对非浮点数的算术运算会产生一个int结果

Answer 3

    private void shiftAndCheck(final byte value, final int shift) {
        byte result = (byte) (value << shift);
    }

As @Konstantin already said it returns int , so either cast it into byte or use int result .