I think Perl does both: taking $i%32 as its RHS, and drop the higher bits, just keeping the lower 32 bits, from the result.
No, perl is doing exactly as it says in the docs; it is just using the underlying C << operator, whose result is undefined for overflows. That means the result will vary based on C compiler and/or CPU architecture.
from pp.c:
const IV shift = POPi;
if (PL_op->op_private & HINT_INTEGER) {
const IV i = TOPi;
SETi(i << shift);
}
else {
const UV u = TOPu;
SETu(u << shift);
}
Dave.
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