[kazinator]

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.

[rcfox]

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

[Findecanor]

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.

[kazinator]

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.