64 bit integer

rajesh_b · 07-18-2006, 04:44 AM

hi,
In one of my structure one field is of 64 bit width. I know that i can use "long long" data type of linux, but it is linux extension. And i want my program to be portable. Can any body suggest me how can i declare and manipulate that field. Thanks in advance.

regards
Rajesh

dmail · 07-18-2006, 04:53 AM

"long long" does not guarantee a 64bit int.

you should be able to include stdint.h on linux and mac
which has datatypes.
I havn't come across too many windows boxs which have this file so I do the following, which may help you.
This is assuming that it is a 64 bit machine and that I have already determinded the box on which it is running.

Code:

#	if ( defined (DE_WINDOWS_BUILD) )
		typedef signed	__int8		int8;
		typedef unsigned __int8		uint8;
		typedef signed	__int16		int16;
		typedef unsigned __int16	uint16;
		typedef signed	__int32		int32;
		typedef unsigned __int32	uint32;
		typedef signed __int64          int64;
		typedef unsigned __int64        uint64;
#	elif (defined(DE_UNIX_BUILD) || defined(DE_MAC_BUILD))
#		include <stdint.h>
		typedef int8_t				int8;
		typedef uint8_t				uint8;
		typedef int16_t				int16;
		typedef uint16_t			uint16;
		typedef int32_t				int32;
		typedef uint32_t			uint32;
		typedef int64_t                         int64;
		typedef uint64_t                        uint64;
#	endif

rajesh_b · 07-18-2006, 05:58 AM

hi dmail,
Thanks for u r reply.
If the processor is of 32-bit type then i think we dont have datatype
uint64.
So what will be the scenario in that case. I hope it is clear.

Regards,
Rajesh.

dmail · 07-18-2006, 06:20 AM

Quote:

If the processor is of 32-bit type then i think we dont have datatype
uint64.

I don't have 64bit machine, but a 32bit gives the output below. I will have you check within windows but I would assume it also defines a 64bit int which is probably a struct with high and low bits.

Code:

#include <iostream>
#include <stdint.h>

int main(int argc, char *argv[])
{
	std::cout <<"int32 :" <<sizeof(int32_t) <<std::endl
	<< "int64 :" <<sizeof(int64_t) <<std::endl;

	return 1;
}

output:
int32 :4
int64 :8

dmail · 07-18-2006, 06:43 AM

Running this on widows (32bit) gives the same results so I can't see there being a problem

Code:

#include <iostream>

int main(int argc, char *argv[])
{
	std::cout <<"int32 :" <<sizeof(__int32) <<std::endl
	<< "int64 :" <<sizeof(__int64) <<std::endl;
	return 1;
}

output:
int32 :4
int64 :8