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From: Keith Thompson <kst-u@mib.org>
Newsgroups: comp.lang.c
Subject: Re: C wish list
Date: Wed, 20 Mar 2019 17:52:01 -0700
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Ben Bacarisse <ben.usenet@bsb.me.uk> writes:
> Keith Thompson <kst-u@mib.org> writes:
[...]
>> The proposed wording is:
>>
>>     The value of E1 << E2 is E1 left-shifted E2 bit positions; vacated
>>     bits are zero-filled. The value of the result is E1×2^E2, reduced
>>     modulo 2^N, where N is the number of value bits in the result
>>     type.
>
> Doesn't that make 1 << 31 yield 0 for 32-bit signed int?  The results
> would also be rather "unexpected" for smaller shifts of negative numbers
> (e.g. -1 << 0 would be 2147483647 AKA 0x7fffffff).
>
> <cut>

Yes, it does.  I made some lazy and incorrect assumptions.

I think the intent is that the << operator should operate on bits in the
same way for signed and unsigned types.  For 32-bit unsigned int, 1U<<31
yields 2147483648.  That bit pattern represents -2147483648 in 32-bit
signed int, but the wording I quoted would reduce that modulo
2147483648, yielding 0.

Using gcc on my system, 1<<31 yields -2147483648 (though the behavior is
undefined).

Either the proposal is inconsistent (perhaps just in this case?) or I'm
misreading it.

-- 
Keith Thompson (The_Other_Keith) kst-u@mib.org  <http://www.ghoti.net/~kst>
Will write code for food.
void Void(void) { Void(); } /* The recursive call of the void */