Sizeof(X) on different architectures

[Martin Roos]

hy ppl, i'm trying to create some multiplatformed software here and i'm very curious about the sizes of common variables on different machines.

if you are running something different than a 32-bit x86 box under your desk, please check what this program outputs.

//-- program starts --

#include <stdio.h>

#define showsize(x,y) printf("Size of %s %d \n",x,sizeof(y));

typedef struct db_s{
int offset;
int length;
char* data;
} db;

int main() {
showsize("char",char);
showsize("int",int);
showsize("long",long);
showsize("char*",char*);
showsize("int*",int*);
showsize("long*",long*);
showsize("void*",void*);
showsize("db",db);
showsize("db*",db*);
return 0;
}

//-- program ends --

on an 32bit x86 it reports like this :

Size of char 1
Size of int 4
Size of long 4
Size of char* 4
Size of int* 4
Size of long* 4
Size of void* 4
Size of db 12
Size of db* 4


......

are there any 64bit sun users out there ? 64bit x86 machine owners and mac users ?

if you can please report in

 

[Martin Roos]

some c compilers seem to have issues with the #define line,
so here is a defineless version

#include <stdio.h>


typedef struct db_s{
int offset;
int length;
char* data;
} db;

void report(char *name, int size) {
printf("Size of %s %d \n",name,size);
}

int main() {
report("char",sizeof(char));
report("int",sizeof(int));
report("long",sizeof(long));
report("char*",sizeof(char*));
report("int*",sizeof(int*));
report("long*",sizeof(long*));
report("void*",sizeof(void*));
report("db",sizeof(db));
report("db*",sizeof(db*));
return 0;
}

 

[pete]

#define showsize(x,y) printf("Size of %s %d \n",x,sizeof(y));



Size of char 1

sizeof(char) is always one.

Size of char* 4

Size of void* 4

sizeof(char*) is equal to sizeof(void*),
as long as they're on the same implementation.

 

[Jonathan Adams]

hy ppl, i'm trying to create some multiplatformed software here and i'm very
curious about the sizes of common variables on different machines.

if you are running something different than a 32-bit x86 box under your desk,
please check what this program outputs.

//-- program starts --

#include <stdio.h>

#define showsize(x,y) printf("Size of %s %d \n",x,sizeof(y));

typedef struct db_s{
int offset;
int length;
char* data;
} db;
....

on an 32bit x86 it reports like this :

Size of char 1
Size of int 4
Size of long 4
Size of char* 4
Size of int* 4
Size of long* 4
Size of void* 4
Size of db 12
Size of db* 4

32-bit sparc is identical. 64-bit sparc, and all recent 64-bit Unixes
that I know of use the "LP64" model -- longs and pointers are 64-bit,
ints are 32-bit. I think Windows 64-bit uses the (crazy, IMHO) "LLP64"
model -- long longs and pointers are 64-bits, longs and ints are 32-bit.

The typical Unix and Windows 32-bit case has been referred to as "ILP32"
(ints, longs, and pointers are 32-bits).

In any case, the output when compiled 64-bit on Solaris sparc is:

Size of char 1
Size of int 4
Size of long 8
Size of char * 8
Size of int * 8
Size of long * 8
Size of void * 8
Size of db 16
Size of db * 8

Cheers,
- jonathan

 

[Kenneth Brody]

hy ppl, i'm trying to create some multiplatformed software here and i'm
very curious about the sizes of common variables on different machines.

[...]

On a Cray Y/MP EL under UNICOS 9.0:

Size of char 1
Size of int 8
Size of long 8
Size of char* 8
Size of int* 8
Size of long* 8
Size of void* 8
Size of db 24
Size of db* 8

 

[Kurt Watzka]

sizeof(void*) is undefined in standard C.

Why do you think so? While "void" is an incomplete type
that cannot be completed, "void *" is a pointer type, and
it is not incomplete.

Kurt Watzka

 

[CBFalconer]

.... snip ...

sizeof(char*) is equal to sizeof(void*),
as long as they're on the same implementation.

sizeof(void*) is undefined in standard C.

--
Some useful references:
<http://www.ungerhu.com/jxh/clc.welcome.txt>
<http://www.eskimo.com/~scs/C-faq/top.html>
<http://benpfaff.org/writings/clc/off-topic.html>
<http://anubis.dkuug.dk/jtc1/sc22/wg14/www/docs/n869/> (C99)
<http://www.dinkumware.com/refxc.html> C-library

 

[Goran Larsson]

sizeof(void*) is undefined in standard C.

How can we have a void* with the same representation (and interchangeability)
as char* [ISO/IEC9899:1999 6.2.5 §26] and at the same time have sizeof(void*)
undefined?

A void* must have the same size as char*. The void type, on the other hand,
doesn't have a size so sizeof(void) is undefined.

 

[Derrick Coetzee]

#define showsize(x,y) printf("Size of %s %d \n",x,sizeof(y));

A slightly cleverer way of doing this is with stringizing:

#define showsize(t) printf("Size of %s %d \n", #t, sizeof(t));

 

[Keith Thompson]

some c compilers seem to have issues with the #define line,
so here is a defineless version

#include <stdio.h>


typedef struct db_s{
int offset;
int length;
char* data;
} db;

void report(char *name, int size) {
printf("Size of %s %d \n",name,size);
}

int main() {
report("char",sizeof(char));
report("int",sizeof(int));
report("long",sizeof(long));
report("char*",sizeof(char*));
report("int*",sizeof(int*));
report("long*",sizeof(long*));
report("void*",sizeof(void*));
report("db",sizeof(db));
report("db*",sizeof(db*));
return 0;
}

I'll use the system identification strings reported by the GNU
"config.guess" script.

mips-sgi-irix6.5
Size of char 1
Size of int 4
Size of long 4
Size of char* 4
Size of int* 4
Size of long* 4
Size of void* 4
Size of db 12
Size of db* 4

ia64-unknown-linux-gnu:
Size of char 1
Size of int 4
Size of long 8
Size of char* 8
Size of int* 8
Size of long* 8
Size of void* 8
Size of db 16
Size of db* 8

alphaev68-dec-osf5.1b:
Size of char 1
Size of int 4
Size of long 8
Size of char* 8
Size of int* 8
Size of long* 8
Size of void* 8
Size of db 16
Size of db* 8

sv1-cray-unicos10.0.1.X:
Size of char 1
Size of int 8
Size of long 8
Size of char* 8
Size of int* 8
Size of long* 8
Size of void* 8
Size of db 24
Size of db* 8

powerpc-ibm-aix5.2.0.0:
Size of char 1
Size of int 4
Size of long 4
Size of char* 4
Size of int* 4
Size of long* 4
Size of void* 4
Size of db 12
Size of db* 4

VAX/VMS (no config.guess script):
Size of char 1
Size of int 4
Size of long 4
Size of char* 4
Size of int* 4
Size of long* 4
Size of void* 4
Size of db 12
Size of db* 4

 

[CBFalconer]

Why do you think so? While "void" is an incomplete type
that cannot be completed, "void *" is a pointer type, and
it is not incomplete.

Woops. I was overhasty. I must have been thinking of *voidptr.

 

[Neil Kurzman]

You realize those sizes come from the compiler, not the CPU.
compile the code with a 16bit windows compiler and it will report the same numbers in 16,32, and 64 bit systems.
The sizes are listed in one of the header files

 

[Dan Pop]

You realize those sizes come from the compiler, not the CPU.

In most cases, the implementor's chices are *strongly* influenced by the
CPU. In some cases, they are also influenced by the OS.

Dan

 

[Dan Pop]

sizeof(void*) is undefined in standard C.

Engage your brain and explain why.

Dan

 

[Dan Pop]

A slightly cleverer way of doing this is with stringizing:

#define showsize(t) printf("Size of %s %d \n", #t, sizeof(t));

^^ ^^^^^^^^^
Your time would have been better spent in removing the undefined
behaviour: cleverness is no substitute for correctness.

Dan

 

[Keith Thompson]

Engage your brain and explain why.

CBFalconer acknowledged his mistake two days ago:

] Woops. I was overhasty. I must have been thinking of *voidptr.

 

[Dan Pop]

Engage your brain and explain why.

CBFalconer acknowledged his mistake two days ago:

] Woops. I was overhasty. I must have been thinking of *voidptr.

I got that acknowledgment, in the meantime, too.

Dan