You don't explain it. If you take the lower 32 bits of the result that's too big to fit into an integer, I expect the higher bits to be thrown away, and the lower bits kept. But the lower 32 bits of 4294967296 is 0, not 1.

I think Zaxo's reply is much closer to the truth.

If I do 

$\ = "\n";
print 2147483648  << 1;

[download]
then I get 0. Even more: 
print +(2147483648+1) << 1;

[download]
yields 2.

I think Perl does both (edit: at least, on my platform): taking $i%32 as its RHS, and drop the higher bits, just keeping the lower 32 bits, from the result.

I just wish jeshuashok's loop counter went a bit higher, to at least above 64.

In reply to Re^2: Left shift operation done more than 32 times by bart
in thread Left shift operation done more than 32 times by jesuashok

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