Arithmetic bit-shift on a signed integer

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Right shift of a negative signed number has implementation-defined behaviour.

If your 8 bits are meant to represent a signed 8 bit value (as you’re talking about a “signed 32 bit integer” before switching to 8 bit examples) then you have a negative number. Shifting it right may fill “empty” bits with the original MSB (i.e. perform sign extension) or it may shift in zeroes, depending on platform and/or compiler.

(Implementation-defined behaviour means that the compiler will do something sensible, but in a platform-dependent manner; the compiler documentation is supposed to tell you what.)

A left shift, if the number either starts out negative, or the shift operation would shift a 1 either to or beyond the sign bit, has undefined behaviour (as do most operations on signed values which cause an overflow).

(Undefined behaviour means that anything at all could happen.)

The same operations on unsigned values are well-defined in both cases: the “empty” bits will be filled with 0.

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