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sulhaq

macrumors regular
Jul 29, 2006
198
0
Significant Figure conversion error and Rounding off error. 1 Kilo does not mean 1000 in the computer world it means 1024. If you look in system information 2.33ghz translates to 2327 Mhz. Similarly due to this error the total RAM is 1.98gb which windows shows correctly. OSX just rounds it off to 2gb and doesn't care about conversion errors.
 

daneoni

macrumors G4
Original poster
Mar 24, 2006
11,832
1,564
It also seems the merom chip can support 977MHz of FSB. So 677 must be very limiting but its better than 167Mhz...ugh!
 

daneoni

macrumors G4
Original poster
Mar 24, 2006
11,832
1,564
Significant Figure conversion error and Rounding off error. 1 Kilo does not mean 1000 in the computer world it means 1024. If you look in system information 2.33ghz translates to 2327 Mhz. Similarly due to this error the total RAM is 1.98gb which windows shows correctly. OSX just rounds it off to 2gb and doesn't care about conversion errors.

I see
 

PlaceofDis

macrumors Core
Jan 6, 2004
19,241
6
Significant Figure conversion error and Rounding off error. 1 Kilo does not mean 1000 in the computer world it means 1024. If you look in system information 2.33ghz translates to 2327 Mhz. Similarly due to this error the total RAM is 1.98gb which windows shows correctly. OSX just rounds it off to 2gb and doesn't care about conversion errors.

um, no? 1024 + 1024 = 2048 technically then, not 1.98 no?
 

gekko513

macrumors 603
Oct 16, 2003
6,301
1
We're talking about RAM here, right? A quick google search tells me that this is common in Windows XP and isn't related to the MacBook (Pro) in particular. I haven't found why, yet.
 

drval

macrumors member
Oct 19, 2006
53
0
This is really out of the dusty halls of long forgotten trivia but, yes, it's 2048. There is always a 64K core kernel so take that off and now your're at:

2048
- 64
____
1984

So with decimal rounding/truncation, you get 1.98(4) GB.

So, yes, Windows is "more" precise in this regard but it may well be that Mac has a differet kernel allocation algorithm -- so that 2(.000) GB might actually be right!
 

PlaceofDis

macrumors Core
Jan 6, 2004
19,241
6
This is really out of the dusty halls of long forgotten trivia but, yes, it's 2048. There is always a 64K core kernel so take that off and now your're at:

2048
- 64
____
1984

So with decimal rounding/truncation, you get 1.98(4) GB.

So, yes, Windows is "more" precise in this regard but it may well be that Mac has a differet kernel allocation algorithm -- so that 2(.000) GB might actually be right!

thank you for that. honestly. its always nice to know the specifics of these things than to leave them up in the air. :D
 

drval

macrumors member
Oct 19, 2006
53
0
I believe that has to do with how 32-bit Windows handles "blocks" of RAM. The largest addressable block is "2GB" so, having more doesn't give you an overall "larger" block as far as that goes; but it does give you almost a second such block.

In otherwords, accessing any address space over 2 GB will be handled more like a memory swap than anything else.

This means that all that the 32-bit Windows kernel "sees" are blocks up to 2 GB.
 
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