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const and 'constant expression'

Hi C experts

I've a bunch of questions.

Consider this simplified piece of code.

const int a = 10;

int main () {
static int b = a;
return b;
}

~>gcc test.c
test.c: In function `main':
test.c:4: error: initializer element is not constant
~>

Question1
I understand that const is 'read-only' but I could not convince my
co-worker that its value can change.

6.7.8 Initialization

[...]

[#4] All the expressions in an initializer for an object
that has static storage duration shall be constant
expressions or string literals.

Can someone point me to text that says that 'const' is not 'constant
expression' and can in fact change?

He claims that the compiler may choose to substitute the value at
compile time.
I tried to explain how the value can change (but I had to use pointers
to const int) which he shot down saying the comparison of examples is
not valid (it's not initializer and it involves pointers).

Question2
Moreover, if we turn on optimization, (and assume that the real code
does not have any dead code like above), the compiler does not give any
error. Is the compiler correct after optimisation?
I think it is a bug in compiler but I need text in standard to prove
that.

Thanks in advance for your comments
-Dev

Apr 19 '06 #1
13 2569
de**********@ya hoo.com <de**********@y ahoo.com> a écrit*:
I've a bunch of questions.
Consider this simplified piece of code.

const int a = 10;
int main () {
static int b = a;
return b;
}

~>gcc test.c
test.c: In function `main':
test.c:4: error: initializer element is not constant
~>

Question1
I understand that const is 'read-only' but I could not convince my
co-worker that its value can change.


#include <stdlib.h>
#include <assert.h>
int main(int argc, char *argv []) {
assert( argc > 0 );
const int a= atoi(argv[1]);
static int b= a;
return b;
}

Marc Boyer
Apr 19 '06 #2
de**********@ya hoo.com wrote:
const int a = 10;

int main () {
static int b = a;
return b;
}

~>gcc test.c
test.c: In function `main':
test.c:4: error: initializer element is not constant
~>

Question1
I understand that const is 'read-only' but I could not convince my
co-worker that its value can change.
Whether or not it can is of no import to that error. It is not a
constant, as defined by the Standard. That's all that matters.
<http://c-faq.com/ansi/constasconst.ht ml>.
6.7.8 Initialization

[...]

[#4] All the expressions in an initializer for an object
that has static storage duration shall be constant
expressions or string literals.

Can someone point me to text that says that 'const' is not 'constant
expression' and can in fact change?
Any object, whether const or not, is not a constant expression: 6.6.
"Can in fact change": immaterial.
He claims that the compiler may choose to substitute the value at
compile time.
He may claim whatever he wants. It remains a const object, not a
constant expression, and is therefore not a portable initaliser.
Moreover, if we turn on optimization, (and assume that the real code
does not have any dead code like above), the compiler does not give any
error. Is the compiler correct after optimisation?


By 6.6#10, yes. However, no other compiler (nor even the same compiler
the next time) is required to accept its extensions.

Richard
Apr 19 '06 #3
Thanks Marc, point noted.

Now getting back to my example,

const int a = 10;

In this case, can value of 'a' change? If so how?
If not, why can not variable 'a' be treated a compile time constant.

-Dev

Apr 19 '06 #4
Le 19-04-2006, de**********@ya hoo.com <de**********@y ahoo.com> a écrit*:
Thanks Marc, point noted.

Now getting back to my example,

const int a = 10;

In this case, can value of 'a' change?
Not to my knowledge.
If so how?
If not, why can not variable 'a' be treated a compile time constant.


It 'can' in this case. But, just try to explain in wich case
it can and when it can not....

Marc Boyer
--
Si tu peux supporter d'entendre tes paroles
Travesties par des gueux pour exciter des sots
IF -- Rudyard Kipling (Trad. Paul Éluard)
Apr 19 '06 #5
If the compiler can treat this 'a' as a compile time constant then this
should also work

const int size = 5;
int array[size];

But seems like it does not compile even with optimizations.

This is causing confusion.

-Dev

Apr 19 '06 #6
Marc Boyer wrote:

Le 19-04-2006, de**********@ya hoo.com <de**********@y ahoo.com> a écrit :
Thanks Marc, point noted.

Now getting back to my example,

const int a = 10;

In this case, can value of 'a' change?


Not to my knowledge.


Isn't this legal?

==========

#include <stdio.h>

const int a = 10;

int main(int argc,char *argv[])
{
int *pt = (int *)&a;

printf("%d\n",a );

*pt = 20;

printf("%d\n",a );
}

==========

It fails here, at runtime, but only because "a" has been placed into
read-only memory on this platform. But it's perfectly "legal" C, as
far as I know. It compiles without even a warning about "pt" and "a",
even with maximum warnings turned on.

[...]

--
+-------------------------+--------------------+-----------------------------+
| Kenneth J. Brody | www.hvcomputer.com | |
| kenbrody/at\spamcop.net | www.fptech.com | #include <std_disclaimer .h> |
+-------------------------+--------------------+-----------------------------+
Don't e-mail me at: <mailto:Th***** ********@gmail. com>
Apr 19 '06 #7
de**********@ya hoo.com wrote:
If the compiler can treat this 'a' as a compile time constant then
this should also work


Please read the information below.

Brian

--
Please quote enough of the previous message for context. To do so from
Google, click "show options" and use the Reply shown in the expanded
header.
Apr 19 '06 #8
Kenneth Brody wrote:
Isn't this legal?

==========

#include <stdio.h>

const int a = 10; int *pt = (int *)&a; It fails here, at runtime, but only because "a" has been placed into
read-only memory on this platform. But it's perfectly "legal" C, as
far as I know. It compiles without even a warning about "pt" and "a",
even with maximum warnings turned on.


What do you mean by "legal"? It doesn't require a diagnostic, but it's
undefined behavior.

Brian
Apr 19 '06 #9
de**********@ya hoo.com wrote:

Please provide context when replying. Usenet is not Google Groups,
Google Groups is merely on of many Usenet service provides that provides
a poor interface. There is no guarantee that others have (or ever will)
see the post you are replying to.
If the compiler can treat this 'a' as a compile time constant then this
should also work

const int size = 5;
int array[size];

But seems like it does not compile even with optimizations.

This is causing confusion.


Basically, you can't do it because the standard says you can't. const in
C is a promise that you will not attempt to modify something, it does
not declare a real constant. So, for example, you can do:
volatile const int fred;
Then, you might be able to use some magic in your linker to map fred
over a hardware register and each time you read it you will get the
current contents of that register.

Note that C99 added variable length arrays, so in C99 you could do:
int main(void)
{
const int size = 5;
int array[size];
/* do stuff */
return 0;
}

However, in the above array is a variable length array where the length
is determined at run time, not a normal fixed length array.
--
Flash Gordon, living in interesting times.
Web site - http://home.flash-gordon.me.uk/
comp.lang.c posting guidelines and intro:
http://clc-wiki.net/wiki/Intro_to_clc
Apr 19 '06 #10

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