«

My code obfuscator gave me this:

char buff[1, 9];

to which gcc retorted:

"ISO C90 forbids variable-size array 'buff'"

and checking the standard, it appears that commas are indeed forbidden
from being in a constant expression.

Why's that? Wouldn't it make more sense to allow commas as long as the
left and right operands are also constant expressions? Calling the
expression 1,9 "variable-size" is pretty silly.

-Peter

In article <news:clpci2$hkg$>
Peter Ammon <> wrote:
>... checking the standard, it appears that commas are indeed forbidden
>from being in a constant expression.
>
>Why's that? Wouldn't it make more sense to allow commas as long as the
>left and right operands are also constant expressions? Calling the
>expression 1,9 "variable-size" is pretty silly.

Asking for sense from the Standards Committees may be going a bit
overboard.

Seriously, even if the left operand is *not* constant, the
expression itself could still be a constant. For instance:

i = (printf("hello\n"), 2);

is certain to set i to 2. In this case, however, there is also
a required side-effect (writing "hello\n" to stdout). As such,
the parenthesized expression is clearly unsuitable for use as a
static initializer:

static int x = (printf("hello\n"), 2); /* WRONG */

but might be allowed as a constant in "more general" positions:

printf("I say ");
{
int i, x[(printf("hello\n"), 2)];
for (i = 0; i < sizeof x / sizeof *x; i++)
x[i] = printf("Hello hello\n");
printf("I don't know why you say goodbye\n");
printf("I say hello\n");
}

So perhaps, once one starts down this road, a committee subgroup
might not be able to agree as to precisely what is allowed when.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.

Peter Ammon wrote:
> My code obfuscator gave me this:
>
> char buff[1, 9];
>
> to which gcc retorted:
>
> "ISO C90 forbids variable-size array 'buff'"
>
> and checking the standard, it appears that commas are indeed forbidden
> from being in a constant expression.
>
> Why's that? Wouldn't it make more sense to allow commas as long as the
> left and right operands are also constant expressions? Calling the
> expression 1,9 "variable-size" is pretty silly.
>

"1, 9" is an *expression* in C, not a constant.

If what you're trying to do is declare a multidimensional array, that
would be:

char buff[1][9];

Please look this up in whatever C book you have -- *and* (as you should
have done *before* posting) read the FAQ.

http://www.eskimo.com/~scs/C-faq/top.html

HTH,
--ag
--
Artie Gold -- Austin, Texas

"If you don't think it matters, you're not paying attention."

Peter Ammon wrote:

> My code obfuscator gave me this:
>
> char buff[1, 9];
>
> to which gcc retorted:
>
> "ISO C90 forbids variable-size array 'buff'"
>
> and checking the standard, it appears that
> commas are indeed forbidden from being in a constant expression.
>
> Why's that? Wouldn't it make more sense to allow commas
> as long as the left and right operands are also constant expressions?
> Calling the expression 1,9 "variable-size" is pretty silly.

> cat buff.c
char buff[1, 9];

> gcc -Wall -std=c99 -pedantic -c buff.c
buff.c:1: error: parse error before ',' token
> gcc --version
gcc (GCC) 3.4.2

Peter Ammon <> wrote in message news:<clpci2$hkg$>...
> My code obfuscator gave me this:
>
> char buff[1, 9];
>
> to which gcc retorted:
>
> "ISO C90 forbids variable-size array 'buff'"
>
> and checking the standard, it appears that commas are indeed forbidden
> from being in a constant expression.
>
> Why's that? Wouldn't it make more sense to allow commas as long as the
> left and right operands are also constant expressions? Calling the
> expression 1,9 "variable-size" is pretty silly.
>
> -Peter

In C the size of an array must be constant. 1,9 is not a constant.

Artie Gold wrote:
> Peter Ammon wrote:
>
>> My code obfuscator gave me this:
>>
>> char buff[1, 9];
>>
>> to which gcc retorted:
>>
>> "ISO C90 forbids variable-size array 'buff'"
>>
>> and checking the standard, it appears that commas are indeed forbidden
>> from being in a constant expression.
>>
>> Why's that? Wouldn't it make more sense to allow commas as long as
>> the left and right operands are also constant expressions? Calling
>> the expression 1,9 "variable-size" is pretty silly.
>>
>
> "1, 9" is an *expression* in C, not a constant.

Thank you for this correction. What I really intended to say was
"constant expression," not constant.

>
> If what you're trying to do is declare a multidimensional array, that
> would be:
>
> char buff[1][9];
>
> Please look this up in whatever C book you have -- *and* (as you should
> have done *before* posting) read the FAQ.
>
> http://www.eskimo.com/~scs/C-faq/top.html

I cannot find any discussion of why expressions with commas cannot be
constant expressions in K&R, C Unleashed, or the FAQ.

-Peter


--
Pull out a splinter to reply.

E. Robert Tisdale wrote:
> Peter Ammon wrote:
>
>> My code obfuscator gave me this:
>>
>> char buff[1, 9];
>>
>> to which gcc retorted:
>>
>> "ISO C90 forbids variable-size array 'buff'"
>>
>> and checking the standard, it appears that
>> commas are indeed forbidden from being in a constant expression.
>>
>> Why's that? Wouldn't it make more sense to allow commas
>> as long as the left and right operands are also constant expressions?
>> Calling the expression 1,9 "variable-size" is pretty silly.
>
>
> > cat buff.c
> char buff[1, 9];
>
> > gcc -Wall -std=c99 -pedantic -c buff.c
> buff.c:1: error: parse error before ',' token
> > gcc --version
> gcc (GCC) 3.4.2

Curious regression; the error message emitted by my version (gcc 3.3)
seems much more informative.

--
Pull out a splinter to reply.

Stuart Gerchick wrote:
> Peter Ammon <> wrote in message news:<clpci2$hkg$>...
>
>>My code obfuscator gave me this:
>>
>>char buff[1, 9];
>>
>>to which gcc retorted:
>>
>>"ISO C90 forbids variable-size array 'buff'"
>>
>>and checking the standard, it appears that commas are indeed forbidden
>>from being in a constant expression.
>>
>>Why's that? Wouldn't it make more sense to allow commas as long as the
>>left and right operands are also constant expressions? Calling the
>>expression 1,9 "variable-size" is pretty silly.
>>
>>-Peter
>
>
> In C the size of an array must be constant. 1,9 is not a constant.

1+9 is a constant expression. Why not have 1,9 a constant expression as
well?

--
Pull out a splinter to reply.

"Peter Ammon" <> wrote in message
news:%h3gd.36079$. com...
> Stuart Gerchick wrote:
> > Peter Ammon <> wrote in message
news:<clpci2$hkg$>...
> >
> >>My code obfuscator gave me this:
> >>
> >>char buff[1, 9];
> >>
> >>to which gcc retorted:
> >>
> >>"ISO C90 forbids variable-size array 'buff'"
> >>
> >>and checking the standard, it appears that commas are indeed forbidden
> >>from being in a constant expression.
> >>
> >>Why's that? Wouldn't it make more sense to allow commas as long as the
> >>left and right operands are also constant expressions? Calling the
> >>expression 1,9 "variable-size" is pretty silly.
> >>
> >>-Peter
> >
> >
> > In C the size of an array must be constant. 1,9 is not a constant.
>
> 1+9 is a constant expression. Why not have 1,9 a constant expression as
> well?

To what end? It would invariably evaluate to 9. Hence a comma in a constant
expression would seem a bit weird...

Stuart Gerchick wrote:
> [snip]
> In C the size of an array must be constant. 1,9 is not a constant.
No. See paragraph 6.7.5.2 of the standard: an array that isn't an object
with static storage duration can have a variable length declaration. But
1,9 is not a variable - it is an expression.

Robert