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From: Keith Thompson <kst-u@mib.org>
Newsgroups: comp.lang.c
Subject: Re: Future of C
Date: Tue, 13 Mar 2018 14:26:55 -0700
Organization: None to speak of
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David Brown <david.brown@hesbynett.no> writes:
> On 13/03/18 21:39, Keith Thompson wrote:
>> David Brown <david.brown@hesbynett.no> writes:
>> [...]
>
>>> I am sure there are plenty of details to take care of.  And there are
>>> some oddities with VLA's - if you write "int vs[foo()];", then when you
>>> later write "int n = sizeof(vs);" you have to re-evaluate foo and might
>>> give a result that is not actually the size of "vs".  So you generate
>>> the same code here as if the use had written "int n = 4 * foo();".
>>> Again, this is not /hard/ because you have already done it.
>> 
>> No, evaluating `sizeof vs` does *not* re-evaluate `foo()`.
>
> I'm looking at 6.5.3.4p2 here:
>
> The sizeof operator yields the size (in bytes) of its operand, which
> may be an expression or the parenthesized name of a type. The size is
> determined from the type of the operand. The result is an integer. If
> the type of the operand is a variable length array type, the operand
> is evaluated; otherwise, the operand is not evaluated and the result
> is an integer constant.

Yes.  The type of a VLA object doesn't change after it's been created.

6.7.6.2p5:

    The size of each instance of a variable length array type does
    not change during its lifetime.

The size or length of a VLA type has to be implicitly stored somewhere
for sizeof to work correctly (unless optimization lets the value be
determined at compile time).

-- 
Keith Thompson (The_Other_Keith) kst-u@mib.org  <http://www.ghoti.net/~kst>
Working, but not speaking, for JetHead Development, Inc.
"We must do something.  This is something.  Therefore, we must do this."
    -- Antony Jay and Jonathan Lynn, "Yes Minister"