The overflow warning comes from the fact that you are using "11111111111111111111111111111111" as an octal number. "11111111111111111111111111111111" octal is not a 32-bit number. It's a 96-bit number. Hence the overflow. If you want a 32-bit number, write it as: 0b11111111111111111111111111111111. 
$ perl -wE 'say 0b11111111111111111111111111111111'
4294967295

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In reply to Re: Integer overflow in octal number by JavaFan
in thread Integer overflow in octal number by Anonymous Monk

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