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From: Tim Rentsch <tr.17687@z991.linuxsc.com>
Newsgroups: comp.lang.c
Subject: Re: C wish list
Date: Thu, 21 Mar 2019 22:55:29 -0700
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Keith Thompson <kst-u@mib.org> writes:

> David Brown <david.brown@hesbynett.no> writes:
> [...]
>
>> C has a " << " operator.  It does not have left-shift instructions.  You
>> appear to be mixing up C and assembly again.
>>
>> x << y has the result x * 2^y, reduced by modulo if x is unsigned.  If
>> the value of the result can't fit in the type - which is the case for
>> signed 1 << 31 - then the result is undefined.  C prefers to say it
>> doesn't make sense and the language does not allow it, rather than pick
>> an incorrect and meaningless value for it (regardless of any underlying
>> implementation).
>>
>> So Auntie Spam here is correct.  It is not (AFAIK) the main reason why 1
>> << 31 is not defined (for 32-bit int), but it is a convenient bonus.
>
> 1 << 31 is undefined (assuming 32-bit int) for two reasons.
>
> 6.5.7p3:  It's undefined because the right operand is greater than
> or equal to the width of the promoted right operand.  (0 << 31 is
> undefined for the same reason.)

This part is wrong.  The width of a 32-bit int is 32, not 31.
The /precision/ of a 32-bit int is 31, but it's the width that
matters for the constraints on shifting.  The expression 0 << 31
is well defined for 32-bit int's.