My formula is
_dwSize = ((__dwSize + 4095) >> 12) << 12 ;
is equivalent to (((__dwSize + 4095) / 4096) * 4096)
Suppose __dwSize = 3456 bytes to alloc
(3456 + 4095) = 7551 bytes
7551 / 4096 = 1 (number of pages)
1 * 4096 = 4096 bytes really allocated
For me 4096 is not the same than 7551 or I must return to school. to old...
I proceed like this because I want a quantity representing full memory pages.
But Pelle does not understand
Building Utils.obj.
C:\Users\51966\Documents\#@# Programmation\DIF\Utils.c(111): warning #2287: Dead code removed (around here).
Done
My solution is _dwSize = (__dwSize + 4095) >> 12 ;
_dwSize = _dwSize << 12 ;
I don't like
The project is in Release mode. Is it a bug?
test:#include <stdio.h>
int __cdecl main(void)
{
unsigned long _dwSize, __dwSize;
__dwSize = 3456;
_dwSize = ((__dwSize + 4095) >> 12) << 12 ;
printf("%d\n", _dwSize);
return 0;
}
returns 4096
from podump
1: #include <stdio.h>
2:
3: int __cdecl main(void)
_main:
[00000000] 55 push ebp
[00000001] 89E5 mov ebp,esp
[00000003] 83EC08 sub esp,8
4: {
5: unsigned long _dwSize, __dwSize;
6: __dwSize = 3456;
[00000006] C745F8800D0000 mov dword ptr [ebp-8],D80
7: _dwSize = ((__dwSize + 4095) >> 12) << 12 ;
[0000000D] 8B45F8 mov eax,dword ptr [ebp-8]
[00000010] 8D80FF0F0000 lea eax,[eax+FFF]
[00000016] C1E80C shr eax,C
[00000019] C1E00C shl eax,C
[0000001C] 8945FC mov dword ptr [ebp-4],eax
8: printf("%d\n", _dwSize);
[0000001F] FF75FC push dword ptr [ebp-4]
[00000022] 6800000000 push @9
[00000027] E800000000 call _printf
[0000002C] 83C408 add esp,8
9: return 0;
[0000002F] B800000000 mov eax,0
[00000034] 89EC mov esp,ebp
[00000036] 5D pop ebp
[00000037] C3 ret
1: #include <stdio.h>
2:
3: int __cdecl main(void)
main:
[0000000000000000] 4883EC28 sub rsp,28
4: {
5: unsigned long _dwSize, __dwSize;
6: __dwSize = 3456;
[0000000000000004] C7442420800D0000 mov dword ptr [rsp+20],D80
7: _dwSize = ((__dwSize + 4095) >> 12) << 12 ;
[000000000000000C] 8B442420 mov eax,dword ptr [rsp+20]
[0000000000000010] 678D80FF0F0000 lea eax,[eax+FFF]
[0000000000000017] C1E80C shr eax,C
[000000000000001A] C1E00C shl eax,C
[000000000000001D] 89442424 mov dword ptr [rsp+24],eax
8: printf("%d\n", _dwSize);
[0000000000000021] 488D0D00000000 lea rcx,[@9]
[0000000000000028] 8B542424 mov edx,dword ptr [rsp+24]
[000000000000002C] E800000000 call printf
9: return 0;
[0000000000000031] B800000000 mov eax,0
[0000000000000036] 4883C428 add rsp,28
[000000000000003A] C3 ret
Thank you Timo. I was afraid.
Hi,
Integer math != 'real' math except in limited circumstances
Integer math
dwSize = ((dwSize + 4095) >> 12) << 12 ;
can't be equivalent to the look alike 'real' math
'Real' math
fSize = ((fSize + 4095.0) / 4096) *4096;
To get close you need to preserve at least a few of the decimals from >>12
so
dwSize = ((((dwSize + 4096)*10000) >> 12) << 12)/10000 ; // = 7551
but if need *10000 and then /10000 one might as well use float... unless
no float lib available.
John Z
Ok John but it is slow that is the reason I use shifts
WithQuotedwSize = ((((dwSize + 4096)*10000) >> 12) << 12)/10000 ; // = 7551but if need *10000 and then /1000 one might as well use float... unlessno float lib available
you add another page if the user wants to alloc 4096 byte you give him 8192 bytes
Change with dwSize + 4095
dwSize = ((((dwSize + 4095)*10000) >> 12) << 12)/10000 ; // = 7551
V1 = (dwSize + 4095) ;
V2 = V1 << 10 ;
V3 = V1 << 4 ;
V4 = V1 << 3 ;
V5 = V3 + V4 ;
V6 = V2 - V5 ; ==> (V1 * 1024) - (V1 * 24)
dwSize = ((V6 >> 12) << 12) / 1000 ;ordwSize = (((V1 << 10) - (V1 << 4) - (V1 << 3)) >> 12) << 12) / 1000 ;
:P
Quote from: HellOfMice on November 15, 2024, 12:59:56 PM
you add another page if the user wants to alloc 4096 byte you give him 8192 bytes
Change with dwSize + 4095
dwSize = ((((dwSize + 4095)*10000) >> 12) << 12)/10000 ; // = 7551
oops you are right
Quote from: HellOfMice on November 15, 2024, 12:59:56 PM
V1 = (dwSize + 4095) ;
V2 = V1 << 10 ;
V3 = V1 << 4 ;
V4 = V1 << 3 ;
V5 = V3 + V4 ;
V6 = V2 - V5 ; ==> (V1 * 1024) - (V1 * 24)
dwSize = ((V6 >> 12) << 12) / 1000 ;[/size][/font]
or
dwSize = (((V1 << 10) - (V1 << 4) - (V1 << 3)) >> 12) << 12) / 1000 ;
Very innovative!
John Z
I like to be an original
Quote from: HellOfMice on November 16, 2024, 10:51:47 AM
I like to be an original
unique ;)
just to avoid imul from code, but your example confuse normal coder, as in code speed actually wasn't real reason ?
unsigned long dwSize2 = ((dwSize + 4095) / 4096) * 4096;
Hello Timo.
What I use is your formula, the other one was just for the fun.
Here is a link that could interest you [size=78%]https://github.com/WojciechMula/toys (https://github.com/WojciechMula/toys)[/size]
Now my program works well for its first part but I would like to improve avg, mean and stddev so I am searching SSE instruction to speed it up.
Thank you for your message.
Philippe
Quote from: HellOfMice on November 17, 2024, 09:47:46 AM
Hello Timo.
What I use is your formula, the other one was just for the fun.
Here is a link that could interest you https://github.com/WojciechMula/toys (https://github.com/WojciechMula/toys)
Now my program works well for its first part but I would like to improve avg, mean and stddev so I am searching SSE instruction to speed it up.
Thank you for your message.
Philippe
Similar discussion here :
https://stackoverflow.com/questions/22970621/aligning-virtual-address-to-immediate-next-page-boundary
Many words for doing what Timo does in one line
As Vortex hinted, this should tested:#include <stdio.h>
int __cdecl main(void)
{
unsigned long dwSize = 3456;
unsigned long dwSize2;
for (dwSize = 4090; dwSize < 4100; dwSize++) {
dwSize2 = ((dwSize + 4095) & 0x0000F000);
printf("%d %d\n",dwSize, dwSize2);
}
for (dwSize = 0xFFF5; dwSize < 0x10005; dwSize++) {
dwSize2 = ((dwSize + 4095) & 0xFFFFF000);
printf("%d %d\n",dwSize, dwSize2);
}
return 0;
}
Thank You
I will test
Philippe