Conditional compilation depending on 32-bit or 64-bit architecture

[Gowtham]

Hi, I am trying to write some code which acts differently when
compiled on
32 bit and 64 bit machines. To identify the machine type, I am trying
to find
the sizeof( int ) and comparing it with 32 and 64.

But, the compiler is complaining about syntax errors in the #if lines.

Errors:
32-64.cpp:7:13: missing binary operator before '('
32-64.cpp:12:13: missing binary operator before '('

g++ version: 3.2.3

sizeof() is evaluated at compile time, and hence I am not able to
understand
why this code is not working:

code:

#include <iostream>

using namespace std;

int main( int argc, char *argv[] )
{
#if ( sizeof( int ) == 32 )
cout << "32 bit" << endl;
return 0;
#endif

#if ( sizeof( int ) == 64 )
cout << "64 bit" << endl;
return 0;
#endif

cout << "Neither 32 bit nor 64 bit" << endl;
}

Can somebody please tell me what I have missed?

Thanks
Gowtham

[Gowtham]

On Apr 14, 3:05 pm, Gowtham <[e-mail]:

Hi, I am trying to write some code which acts differently when
compiled on
32 bit and 64 bit machines. To identify the machine type, I am trying
to find
the sizeof( int ) and comparing it with 32 and 64.

But, the compiler is complaining about syntax errors in the #if lines.

Errors:
32-64.cpp:7:13: missing binary operator before '('
32-64.cpp:12:13: missing binary operator before '('

g++ version: 3.2.3

sizeof() is evaluated at compile time, and hence I am not able to
understand
why this code is not working:

code:

#include <iostream>

using namespace std;

int main( int argc, char *argv[] )
{
#if ( sizeof( int ) == 32 )
cout << "32 bit" << endl;
return 0;
#endif

#if ( sizeof( int ) == 64 )
cout << "64 bit" << endl;
return 0;
#endif

cout << "Neither 32 bit nor 64 bit" << endl;

}

Can somebody please tell me what I have missed?

Thanks
Gowtham

Oops, I should have compared sizeof(int) with 4 and 8.
What is the best way of achieving this?

[Sam]

Gowtham writes:

On Apr 14, 3:05 pm, Gowtham <[e-mail]:

>Hi, I am trying to write some code which acts differently when
compiled on
32 bit and 64 bit machines. To identify the machine type, I am trying
to find
the sizeof( int ) and comparing it with 32 and 64.

But, the compiler is complaining about syntax errors in the #if lines.

Errors:
32-64.cpp:7:13: missing binary operator before '('
32-64.cpp:12:13: missing binary operator before '('

g++ version: 3.2.3

sizeof() is evaluated at compile time, and hence I am not able to
understand
why this code is not working:

code:

#include <iostream>

using namespace std;

int main( int argc, char *argv[] )
{
#if ( sizeof( int ) == 32 )
cout << "32 bit" << endl;
return 0;
#endif

#if ( sizeof( int ) == 64 )
cout << "64 bit" << endl;
return 0;
#endif

cout << "Neither 32 bit nor 64 bit" << endl;

}

Can somebody please tell me what I have missed?

Thanks
Gowtham

Oops, I should have compared sizeof(int) with 4 and 8.
What is the best way of achieving this?

The sizeof() operator gets evaluated at compile time, not in the
preprocessor phase, as such it cannot be used in preprocessor directives.

Check the documentation for your compiler or C library. It's fairly likely
that there are some preprocessor macros #define-ed somewhere, that give you
the target platform's bitness.
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[Vladislav.Lazarenko]

Michael,

The version with templates is more elegant, robust and easy to
support.

[Gowtham]

On Apr 14, 4:03 pm, Sam <[e-mail]:

Gowtham writes:

On Apr 14, 3:05 pm, Gowtham <[e-mail]:

Hi, I am trying to write some code which acts differently when
compiled on
32 bit and 64 bit machines. To identify the machine type, I am trying
to find
the sizeof( int ) and comparing it with 32 and 64.

But, the compiler is complaining about syntax errors in the #if lines.

Errors:
32-64.cpp:7:13: missing binary operator before '('
32-64.cpp:12:13: missing binary operator before '('

g++ version: 3.2.3

sizeof() is evaluated at compile time, and hence I am not able to
understand
why this code is not working:

code:

#include <iostream>

using namespace std;

int main( int argc, char *argv[] )
{
#if ( sizeof( int ) == 32 )
cout << "32 bit" << endl;
return 0;
#endif

#if ( sizeof( int ) == 64 )
cout << "64 bit" << endl;
return 0;
#endif

cout << "Neither 32 bit nor 64 bit" << endl;

}

Can somebody please tell me what I have missed?

Thanks
Gowtham

Oops, I should have compared sizeof(int) with 4 and 8.
What is the best way of achieving this?

The sizeof() operator gets evaluated at compile time, not in the
preprocessor phase, as such it cannot be used in preprocessor directives.

Check the documentation for your compiler or C library. It's fairly likely
that there are some preprocessor macros #define-ed somewhere, that give you
the target platform's bitness.

application_pgp-signature_part
1KDownload

Thanks. There is an identifier __WORDSIZE defined in gcc environment.
This will have
values 32 and 64.

I could handle this in my code. Thanks again.

Wondering if this is truly portable?

[=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?=]

On 2008-04-14 12:05, Gowtham wrote:

Hi, I am trying to write some code which acts differently when
compiled on
32 bit and 64 bit machines. To identify the machine type, I am trying
to find
the sizeof( int ) and comparing it with 32 and 64.

But, the compiler is complaining about syntax errors in the #if lines.

Errors:
32-64.cpp:7:13: missing binary operator before '('
32-64.cpp:12:13: missing binary operator before '('

g++ version: 3.2.3

sizeof() is evaluated at compile time, and hence I am not able to
understand
why this code is not working:

code:

#include <iostream>

using namespace std;

int main( int argc, char *argv[] )
{
#if ( sizeof( int ) == 32 )
cout << "32 bit" << endl;
return 0;
#endif

#if ( sizeof( int ) == 64 )
cout << "64 bit" << endl;
return 0;
#endif

cout << "Neither 32 bit nor 64 bit" << endl;
}

Can somebody please tell me what I have missed?

In addition to what others have said I would like to point out that
relying on sizeof(int) is doomed to fail. On many 64-bit platforms the
int is still 32-bits.

--
Erik Wikström

[Sam]

Gowtham writes:

On Apr 14, 4:03 pm, Sam <[e-mail]:

>Check the documentation for your compiler or C library. It's fairly likely
that there are some preprocessor macros #define-ed somewhere, that give you
the target platform's bitness.

application_pgp-signature_part
1KDownload

Thanks. There is an identifier __WORDSIZE defined in gcc environment.
This will have
values 32 and 64.

I could handle this in my code. Thanks again.

Wondering if this is truly portable?

No. If you need to code to work with other compilers, you'll have to
research those too. As someone else pointed out, checking INT_MAX would
probably be more portable.

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[Ian Collins]

Michael DOUBEZ wrote:

>
int main()
{
foo<sizeof(int)>();
return 0;
}

sizeof(void*) would be a better choice. In the two most popular 64 bit
memory models (LP64 and LLP64), sizeof(int) is 4.

--
Ian Collins.

[Andy Champ]

Ian Collins wrote:

Michael DOUBEZ wrote:

>int main()
{
foo<sizeof(int)>();
return 0;
}

sizeof(void*) would be a better choice. In the two most popular 64 bit
memory models (LP64 and LLP64), sizeof(int) is 4.

.... which of course points out that the first question should be:

What do you mean by 64-bit?

(but I like your templates...)

Andy