Weird result of Java Integer left shift

Question

I\'m a little confused now by java left shift operation,

1<<31 =  0x80000000  --> this I can understand

But

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Answer 1

As per the Java Language Specification 15.19. Shift Operators (slightly paraphrased):

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 & with the mask value 0x1f ,or 0b11111. The shift distance actually used is therefore always in the range 0 to 31, inclusive.

That means that (for example) 33, being the 6-bit binary 100001, is reduced to the 5-bit 00001 before being used. So x << 33 is identical to x << 1.

Answer 2

System.out.println(Integer.toBinaryString(1 << 32)); 

Shifts binary 1(10) by 32 times to the left. Hence: 1 in decimal

System.out.println(Integer.toBinaryString(1 << 33)); 

Now, int is of 4 bytes,hence 32 bits. So when you do shift by 33, it's equivalent to shift by 1. Hence : 2 in decimal