From the documentation:
The right shift operator (>>) shifts the first operand the specified number of bits to the right. Excess bits shifted off to the right are discarded. Copies of the leftmost bit are shifted in from the left. Since the new leftmost bit has the same value as the previous leftmost bit, the sign bit (the leftmost bit) does not change. Hence the name "sign-propagating".
From what I understand, since 100 is 0b1100100, shifting it 100 times to the right should yield 0b0. However, when I run 100 >> 100 in Javascript (using chrome), it returns 6. Why is this the case? I am guessing it has something to do with JS's internal representation of numbers but would like to know more clearly.
Edit: The answer is still 6, even when using the unsigned >>> operator. Sign/unsigned does not seem to matter.
Unsigned operation documentation:
The unsigned right shift operator (>>>) (zero-fill right shift) shifts the first operand the specified number of bits to the right. Excess bits shifted off to the right are discarded. Zero bits are shifted in from the left. The sign bit becomes 0, so the result is always non-negative. Unlike the other bitwise operators, zero-fill right shift returns an unsigned 32-bit integer.
The value to the right is taken mod 32. (i.e. Only the last five bits are used.) If you calculate 100>>32, you get 100, which is the same thing you get when you compute 100>>0. After that, 100>>33 is 50, and the cycle repeats.