64x version of Dominions?

[Deathblob]

Originally Posted by Loren

64bit doesn't add one bit of speed unless you're doing 64bit arithmetic. In general it will actually cost you a bit of speed as 64 bit code is bigger than 32 bit code.

Yes! That's so true, if you are reading your 64-bit words over an 8-bit bus.

[Loren]

Originally Posted by Deathblob

Yes! That's so true, if you are reading your 64-bit words over an 8-bit bus.

The memory bus is at least 64 bits, that's not a problem.

The problem is 64 bit code is simply bigger than 32 bit code--fewer lines of code fit into the on-chip caches, you end up with more read cycles to memory.

[Deathblob]

Originally Posted by Loren

The memory bus is at least 64 bits, that's not a problem.

The problem is 64 bit code is simply bigger than 32 bit code--fewer lines of code fit into the on-chip caches, you end up with more read cycles to memory.

Well, I thought those reads should be twice as fast, so I didn't believe you, and went googling.

http://www.osnews.com/story/5768/Are...naries_/page1/

It's a bit old, and Sparc of all things. But looks like you are right.

64bit had ~+20% bigger binaries, ~-10% slower execution.

[Loren]

Originally Posted by Deathblob

Well, I thought those reads should be twice as fast, so I didn't believe you, and went googling.

http://www.osnews.com/story/5768/Are...naries_/page1/

It's a bit old, and Sparc of all things. But looks like you are right.

64bit had ~+20% bigger binaries, ~-10% slower execution.

The 64bit processor reads everything faster, whether the executable is 32bit or 64bit.

[JonBrave]

Originally Posted by Loren

The problem is 64 bit code is simply bigger than 32 bit code--fewer lines of code fit into the on-chip caches, you end up with more read cycles to memory.

How long does it take for the size of "on-chip cache" to double on average PC?