LocalAlloc memory begin

[bomz]

invoke LocalAlloc,LMEM_FIXED OR LMEM_ZEROINIT,N

memory always begin from multiple for 4 point?
as align 4

[hutch--]

There is no difference in 32 bit Windows from GlobalAlloc() using the GMEM_FIXED flag and the GMEM_ZEROINIT flag. In 32 bit windows there is no LOCAL alloc at all, its like many other memory strategies, its mapped to NTDLL internal functions.

[qWord]

You can rely on that the memory returned by Global/LocalAlloc is to be aligned by 8.

[bomz]

the function is not problem I need to be sure that memory allways /4 .
I try 10 times - always /4. but there is no about this in msdn

[qWord]

but there is no about this in msdn

Memory allocated with this function is guaranteed to be aligned on an 8-byte boundary.

GlobalAlloc()

So your on the 'save side' by using this function.

[jj2007]

Here are some steps to produce align 16, in case you want to work with SSE2.

Code Select Expand

include \masm32\include\masm32rt.inc
.686
.xmm

.code
start:
mov ebx, alloc(123) ; produce misalignment ;-)
BytesNeeded=555
mov eax, BytesNeeded
add eax, 16
mov esi, alloc(eax)
print hex$(eax), 9, " result of first alloc", 13, 10
mov edi, esi ; keep that pointer for GlobalFree
add esi, 16 ; add the 16 extra bytes
and esi, -16 ; and align them downwards
movaps xmm0, [esi] ; if it's not aligned 16, you will see a GPF
print hex$(esi), 9, " aligned 16 for use with SSE2", 13, 10
nops 5
free edi
print str$(eax), 9, " result of free edi (0=success)", 13, 10
free ebx
inkey str$(eax), 9, " result of free ebx (0=success)", 13, 10
exit
end start


[bomz]

I think most easy correct begin point from memory begin to 4,8,16...
msdn says that located memory not allways the same size as need - may be greater. the begining is that how windows orginize all memory

[qWord]

I think most easy correct begin point from memory begin to 4,8,16...
msdn says that located memory not allways the same size as need - may be greater. the begining is that how windows orginize all memory

What are you trying to say?

[dedndave]

it doesn't seem to be documented anywhere
so - i align it as desired

1) request (N-1) more bytes than you need, where N = alignment
2) save the original address for the handle to Free the allocated block
3) use the N-aligned address for data
   D = (A+N-1) AND (-N)

A = allocated block address
D = aligned data address

[bomz]

my mathematic english very poor



I mean that windows organize memory using some rule and all part of memory for program, for data, for system, for .... begins from /4
all pages?

[qWord]

bomz,
when allocating memory, there seems to be no documented warranty that the memory is aligned(this what you mean begin/x) in any form. Only GlobalAlloc() guarantees this - it is documented!
(This does not apply to the Virtual memory functions.)

[bomz]

http://msdn.microsoft.com/en-us/library/aa366574%28v=vs.85%29.aspx
where it documented?

[qWord]

follow the link and read the whole text!

[bomz]

Memory allocated with this function is guaranteed to be aligned on an 8-byte boundary. To execute dynamically generated code, use the VirtualAlloc function to allocate memory and the VirtualProtect function to grant PAGE_EXECUTE access.

[MichaelW]

I thought I posted an earlier version of the code in the attachment somewhere here, but I can't seem to find it.