Initially, is
(-1 >>> 0) === (2**32 - 1)However, why does
(-1 >>> 32) === (2**32 - 1)Could it be that
((-1 >>> 31) >>> 1) === 0Performing an unsigned right shift by executing (-1 >>> 0) is crucial here. According to the specification, the result of >>> is always considered as unsigned. When dealing with -1, which is represented in binary as all 1s (e.g., 11111111 in an 8-bit system), we take care to make it unsigned by shifting via >>> 0. This instruction essentially instructs to keep the binary representation of -1 unchanged but to interpret the value as an unsigned number, resulting in a sequence of all 1s.
To corroborate this behavior, one can try specific JavaScript commands in a browser console:
console.log(2**32 - 1) //4294967295
// The '0b' indicates a binary representation, and it can include negativity
console.log(0b11111111111111111111111111111111) //4294967295
console.log(-0b1 >>> 0) //4294967295
An interesting pattern emerges when calculating 2 ** n - 1, where the result will consist of n consecutive ones in binary. For instance, 2**32 - 1 yields 32 ones: 111...111, akin to the exponential relation in binary notation. Similarly, the operation
(-1 >>> 32) === (2**32 - 1)Continuing shifting towards zero bits results in distinctive patterns:
console.log(-1 >>> 31) //1
This clarifies the transition from having primarily zeros to introducing a single one at the far end within a 32-bit configuration.
Further insights are derived through the examination of bitwise operations and their impacts on shifting outcomes per specifications. Exploring scenarios such as -1 >>> 33 unveils the logic behind each bit manipulation step undertaken during shifts. These observations pave the way for comprehensive understanding despite intricate bitwise calculations.
When you perform the operation (-1 >>> 0), you are essentially setting the sign bit to zero while leaving the rest of the number unchanged. This results in the value 2**32 - 1.
The specific behavior described above can be found in the ECMAScript specification. The number of shifts that occur is determined by "masking out all but the least significant 5 bits of rnum, calculated as rnum & 0x1F."
Due to the fact that 32 & 0x1F === 0, both outcomes will be identical.
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