#if constant expression

[kerravon]

According to 6.8 of C90, #if takes a constant expression.

According to 6.4 of C90, the sizeof operator is part of a constant
expression.

Why then do most of my compilers barf on this?

#include <stdio.h>

#if sizeof(int) >= 4
#define XXX "big"
#else
#define XXX "small"
#endif

int main(void)
{
printf("hello %s\n", XXX);
return (0);
}

They barf on the "#if sizeof ..." line.

They explicitly say that I can't use sizeof in a #if

So which bit of the standard did I miss?

BFN. Paul.

[vicks]

#include <stdio.h>

>
#if sizeof(int) >= 4
#define XXX "big"
#else
#define XXX "small"
#endif

int main(void)
{
printf("hello %s\n", XXX);
return (0);

}

[code snnipet...]

So which bit of the standard did I miss?

Well I think that #if can be used only with the Macros(who gets
replaced with contant expressions at compile time) since sizeof gets
resolved at run time I dont think it's good idea to use sizeof with
#if!!

Regards
Vikas Gupta

[James Kuyper]

kerravon wrote:

According to 6.8 of C90, #if takes a constant expression.

According to 6.4 of C90, the sizeof operator is part of a constant
expression.

Why then do most of my compilers barf on this?

#include <stdio.h>

#if sizeof(int) >= 4
#define XXX "big"
#else
#define XXX "small"
#endif

int main(void)
{
printf("hello %s\n", XXX);
return (0);
}

They barf on the "#if sizeof ..." line.

They explicitly say that I can't use sizeof in a #if

So which bit of the standard did I miss?

BFN. Paul.

I don't have a copy of C90; but I suspect that this is not an area where
the differences between C90 and C99 matter. Section 6.10.1p4 of
n1256.pdf says

"After all replacements due to macro expansion and the *defined* unary
operator have been performed, all remaining identifiers (including those
lexically identical to keywords) are replaced with the pp-number 0, "

I believe that this applies to both 'sizeof' and 'int' in your example.
As a result, it becomes:

#if 0(0) >= 4

which explains why it fails.

[Tomás Ó hÉilidhe]

vicks <vi*******@gmai l.comwrote in news:d485a4a6-dd4a-492e-a00b-
9b**********@r1 g2000hsg.google groups.com:

Well I think that #if can be used only with the Macros(who gets
replaced with contant expressions at compile time) since sizeof gets
resolved at run time I dont think it's good idea to use sizeof with
#if!!

sizeof yields a compile-time constant, and is very much _not_ evaluated at
runtime.

--
Tomás Ó hÉilidhe

[pete]

kerravon wrote:

>
According to 6.8 of C90, #if takes a constant expression.

According to 6.4 of C90, the sizeof operator is part of a constant
expression.

The preprocessor can't read keywords.
ISO/IEC 9899: 1990
6.8.1 Conditional inclusion
Constraints

identifiers (including those lexically identical to keywords)
are interpreted as described below;83

83 Because the controlling constant expression
is evaluated during translation phase 4,
all identifiers either are or are not macro names
— there simply are no keywords, enumeration constants, etc.
--
pete

[quarkLore]

On Dec 10, 8:11 am, pete <pfil...@mindsp ring.comwrote:

kerravon wrote:

According to 6.8 of C90, #if takes a constant expression.

According to 6.4 of C90, the sizeof operator is part of a constant
expression.

The preprocessor can't read keywords.

ISO/IEC 9899: 1990
6.8.1 Conditional inclusion
Constraints

identifiers (including those lexically identical to keywords)
are interpreted as described below;83

83 Because the controlling constant expression
is evaluated during translation phase 4,
all identifiers either are or are not macro names
-- there simply are no keywords, enumeration constants, etc.

--
pete

#if is for preprocessor, sizeof is a compile time operator but not
handled at preprocessing stage. If you have gcc put sizeof in your
code and run gcc -E <soure filenameit should give you the
preprocessor output and sizeof would remain as it is.

[Fancois Grieu]

In article
<81************ *************** *******@d21g200 0prf.googlegrou ps.com>,
kerravon <ke******@w3.to wrote:

Why then do most of my compilers barf on this?

#include <stdio.h>
#if sizeof(int) >= 4
#define XXX "big"
#else
#define XXX "small"
#endif
int main(void)
{
printf("hello %s\n", XXX);
return (0);
}

One way to see this is that sizeof is not usable in the proprocessor, which has no
knowledge of types. There is a solution, though

#include <stdio.h>
#define XXX (sizeof(int) >= 4 ? "big" : "small")
int main(void)
{
printf("hello %s\n", XXX);
return (0);
}

If you want to check at compile time that int is big:
/* cause a compile-time error if x is 0 */
#define CompileAssert(x ) do{typedef struct{char f[(x)?1:-1];}assertion_fai lure;}while(0) /* compile time assertion failure */

#include <stdio.h>
int main(void)
{
CompileAssert(s izeof(int) >= 4);
puts("hello big");
return (0);
}
Francois Grieu

[pete]

Keith Thompson wrote:

>
pete <pf*****@mindsp ring.comwrites:

ISO/IEC 9899: 1990
6.8.1 Conditional inclusion
Constraints

identifiers (including those lexically identical to keywords)
are interpreted as described below;83

83 Because the controlling constant expression
is evaluated during translation phase 4,
all identifiers either are or are not macro names
-- there simply are no keywords, enumeration constants, etc.

That quotation is from the C99 standard, not the C90 standard.

That quotation is from the C90 standard.
I think you must be using a public draft instead.
This quotation is from the C99 standard:

ISO/IEC 9899:1999(E)
6.10.1 Conditional inclusion
Constraints

identifiers (including those lexically identical to keywords)
are interpreted as described below;140)

140) Because the controlling constant expression
is evaluated during translation phase 4,
all identifiers either are or are not macro names
— there simply are no keywords, enumeration constants, etc.

--
pete

[Keith Thompson]

pete <pf*****@mindsp ring.comwrites:

Keith Thompson wrote:

>pete <pf*****@mindsp ring.comwrites:

ISO/IEC 9899: 1990
6.8.1 Conditional inclusion
Constraints

identifiers (including those lexically identical to keywords)
are interpreted as described below;83

83 Because the controlling constant expression
is evaluated during translation phase 4,
all identifiers either are or are not macro names
-- there simply are no keywords, enumeration constants, etc.

That quotation is from the C99 standard, not the C90 standard.

That quotation is from the C90 standard.
I think you must be using a public draft instead.

No, I'm using a copy of the actual C90 standard, but you're right, the
quotation is from the C90 standard. I don't know how I missed that.
Apologies.

[...]

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Looking for software development work in the San Diego area.
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Keith Thompson]

pete <pf*****@mindsp ring.comwrites:

Keith Thompson wrote:

>>
pete <pf*****@mindsp ring.comwrites:

ISO/IEC 9899: 1990
6.8.1 Conditional inclusion
Constraints

identifiers (including those lexically identical to keywords)
are interpreted as described below;83

83 Because the controlling constant expression
is evaluated during translation phase 4,
all identifiers either are or are not macro names
-- there simply are no keywords, enumeration constants, etc.

That quotation is from the C99 standard, not the C90 standard.

That quotation is from the C90 standard.
I think you must be using a public draft instead.

No, I'm using a copy of the actual C90 standard, but you're right, the
quotation is from the C90 standard. I don't know how I missed that.
Apologies.

[...]

--
Keith Thompson (The_Other_Keit h) <ks***@mib.or g>
Looking for software development work in the San Diego area.
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"