bit manipulation - How are 32 bit JavaScript numbers resulting from a bit-wise operation converted back to 64 bit numbers

Question

I am trying to understand how bit-wise operation in JavaScript work, more specifically how the 32 bit number resulting from a bit-wise operation is converted back to a 64 bit JavaScript number. I am getting some strange results when setting the left most bit in a 32 bit number and when the operation overflows.

For example, with the following operation:

0x01 << 31

Would normally result in 0x80000000 if the number was 32 bits long. But when JavaScript converts this number back to a 64 bit value, it padds the leftmost 32 bits with 1 resulting in the value FFFFFFFF80000000.

Similarly, when left shifting 32 bits, thus overflowing a 32 bit integer, with the operation:

0x02 << 32

The number would overflow, and the result value should be 0x00. But the resulting JavaScript number is 0x02.

Are there any specific rules that JavaScript uses for bit-wise operation that I am not aware of? I understand that all bit-wise operations are performed with 32 bit integers, and that JavaScript numbers are 64 bit double precision floating point numbers, but I cannot understand where the extra padding comes from when converting between the two.

Answer 1

1. Result of bitwise operators are signed int32's, the sign bit is propagated when they are converted back to Numbers.

2. You cannot shift by more than 31 bits:

Let shiftCount be the result of masking out all but the least significant 5 bits of rnum, that is, compute rnum & 0x1F.

That is, x<<32 is the same as x<<0.