Physical Page Allocator?!

[Perica]

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[Lantz]

Hmm im not sure if this is right, but ill give it a shot. You only need 4 mb of stack if your machine has 4 gb of physical memory, and if it does, 4 mb out of 4 gb isnt much to cry about.

[Perica]

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[uri]

The amount of physical memory isn't likely to change while the PC is running. Simply check how much RAM is installed and set the size of your stack accordingly.

[Slasher]

hi,
lets say that you have 32mb of physical ram and the end of the kernel is marked by k_end.
To find the size of the stack you have to thinks as follows -
each address on the stack is suppose to mark the beginning of a 4kb page that starts on a page boundry.
32mb/4k = 8k entries each of which is 4 bytes(32bits or a long int). So the final size of the stack is 8k entries * 4bytes per entry = 32kb of stack required.
NOTE: 32mb give 32kb stack size,
256mb give 256kb stack size

so you can then do,
stack_start=k_end (stack starts at the end of the kernel)
stack_top=stack_start

to add (push) an address onto the stack
stack_top++;
*stack_top=address

to remove (pop) an address from the stack
address=*stack_top
stack_top--
hope this helps