RE: [Dev-C++] new is a wrapper around malloc so...

[Daniel G. <sp...@ho...>]

Ahhhhhhhhhhh, so that's why everyone wants 64-bits. Games too are requiring 
more, probably because of the same memory-mapping issues.


>From: Per Westermark <pw...@ia...>
>To: Daniel Glenfield <sp...@ho...>
>CC: mu...@gm...,  <dev...@li...>
>Subject: RE: [Dev-C++] new is a wrapper around malloc so...
>Date: Wed, 16 Jul 2003 12:26:07 +0200 (CEST)
>
>A 32-bit pionter is an unsigned entity, so it can span 2^32 bytes - 4GB.
>From this address space, the different OS then reserves areas for OS code,
>buffers, IPC etc. Depending on OS, you may normally not allocate more than
>2 or 3GB for a single application.
>
>This is the big reason why the world is so desperately wanting 64-bit
>machines. In database solutions it is quite common to memory-map the
>database files. However, 2-3GB is a puny database.
>
>/Per W
>
>On Wed, 16 Jul 2003, Daniel Glenfield wrote:
>
> > All I know is that the MSDN documentation I read says 2GB. I'm not sure
> > about how you calculate it all, so you're one up on me!
> >
> >
> > >From: "Abhijit Shylanath" <mu...@gm...>
> > >To: "Daniel Glenfield" <sp...@ho...>
> > >CC: "Dev-Cpp mailing list" <dev...@li...>
> > >Subject: RE: [Dev-C++] new is a wrapper around malloc so...
> > >Date: Wed, 16 Jul 2003 00:40:19 +0530
> > >
> > >
> > > > In 16 bit your code is separated into 'segments' each a maximum of 
>64k
> > >in
> > > > size. 32-bit can have up to 2GB segments (!!!).
> > >
> > >4G, right? (2^32).
> > >
> > >However, that's true only on a 32-bit system (as you've said)... what
> > >platform are you building on? On 16-bit (`real-mode') DOS, even though
> > >sizeof(farptr) will give you four, it cannot address 32-bits worth of
> > >memory. Far pointers work like this:
> > >
> > >	32 bits of farptr  =
> > >	16 bits of segment .
> > >	16 bits of offset
> > >
> > >The effective address is actually calculated by multiplying the segment 
>by
> > >16 decimal (0x10 (which is equivalent to left-shifting by four)) and 
>then
> > >ADDING the offset... so that if your far-ptr has this value:
> > >
> > >	0x1234abcd
> > >
> > >Written as:
> > >
> > >	1234:abcd
> > >
> > >Its effective address, which is worth 20-bits (and can therefore 
>address
> > >2^20 = 1M) is calculated  as:
> > >
> > >	0x12340
> > >+	 0xabcd
> > >-------------
> > >=	0x1cf0d
> > >
> > >Whereas, in the flat memory model(protected-mode), the effective 
>address
> > >would itself be
> > >
> > >	0x1234abcd
> > >
> > >--
> > >Abhijit Shylanath
> > >Homepage: http://mudeth.tripod.com/
> > >
> >
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