Why does Java mask shift operands with 0x1F?

Question

In Java:

(0xFFFFFFFF <<  1) = 0xFFFFFFFE = 0b1111111111111110
                :         :               :
(0xFFFFFFFF << 30) = 0xE0000000 = 0b111         


        

Answer 1

JLS 15.19 If the promoted type of the left-hand operand is int, only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x1f (0b11111). The shift distance actually used is therefore always in the range 0 to 31, inclusive

to put it simply 0xFFFFFFFF << 32 is equivalnt to 0xFFFFFFFF << (32 & 0x1f) is equivalent to 0xFFFFFFFF << 0