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0 pointskazinator2y ago0 comments

However, the oposite shift is undefined if a 1 goes into the sign bit.

More precisely, regarding E1 << E2, it is written (in the April 2023 ISO C draft):

"If E1 has a signed type and nonnegative value, and E1 × 2ᴱ² is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined."

Thus if E1 is negative, or if the result overflows, UB.

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rcfox2y ago

Didn't they specify two's complement signed integers somewhat recently? What's the rationale for leaving these behaviours undefined?

Findecanor2y ago

On some CPUs architectures, the operand size for some instructions could be larger than an `int`, in which case the upper part of a CPU register would become invalid on overflow instead of containing an extension of the sign bit.

There are also CPUs that do "saturating arithmetic" where overflow results in INT_MAX instead of wrapping.

kazinatorOP2y ago

Because the shift operations are arithmetically defined, and the situations that are undefined correspond to overflow. v << 1 means the same thing as v * 2 and is undefined if v * 2 is undefined.

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