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);
    }

[download]

Dave.

In reply to Re^3: Left shift operation done more than 32 times by dave_the_m
in thread Left shift operation done more than 32 times by jesuashok

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