If the shifted number is positive >>> and >> work the same.
If the shifted number is negative >>> fills the most significant bits with 1s whereas >> operation shifts filling the MSBs with 0.
Is my understanding correct?
If the negative numbers are stored with the MSB set to 1 and not the 2s complement way that Java uses the the operators would behave entirely differently, correct?
The way negative numbers are represented is called 2's complement. To demonstrate how this works, take -12 as an example. 12, in binary, is 00001100 (assume integers are 8 bits though in reality they are much bigger). Take the 2's complement by simply inverting every bit, and you get 11110011. Then, simply add 1 to get 11110100. Notice that if you apply the same steps again, you get positive 12 back.
The >>> shifts in zero no matter what, so 12 >>> 1 should give you 00000110, which is 6, and (-12) >>> 1 should give you 01111010, which is 122. If you actually try this in Java, you'll get a much bigger number since Java ints are actually much bigger than 8 bits.
The >> shifts in a bit identical to the highest bit, so that positive numbers stay positive and negative numbers stay negative. 12 >> 1 is 00000110 (still 6) and (-12) >> 1 would be 11111010 which is negative 6.
Definition of the >>> operator in the Java Language Specification:
The value of
n>>>sis n right-shifted s bit positions with zero-extension. If n is positive, then the result is the same as that ofn>>s; if n is negative, the result is equal to that of the expression(n>>s)+(2<<~s)if the type of the left-hand operand isint, and to the result of the expression(n>>s)+(2L<<~s) if the type of the left-hand operand islong.
Just the opposite, the >>> fills with zeros while >> fills with ones if the h.o bit is 1.
