Create buffer divisible by 4

[dedndave]

 :dazzled:

[dedndave]

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align2 proc len:dword,multiple_of_2:dword
;to do: need check if multiple of 2 is so high to fit in this dword

        mov     edx,multiple_of_2
        mov     eax,len
        add     eax,edx
        dec     eax
        neg     edx
        and     eax,edx
        ret

align2 endp

[mineiro]

Fatality, Sr Dedndave wins, flawers victory hehehe.  :thumbu

[drizz]

or

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mov eax,Value
mov edx,Alignment
dec eax
dec edx
or eax,edx
inc eax

and aligned memory alloc

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if @WordSize ne 8
rax equ <eax>
rdx equ <edx>
size_t typedef dword
else
size_t typedef qword
endif

_alloc:
mov rax,_ALLOC_BYTES
add rax,_MEMORY_ALIGN
invoke HeapAlloc,_HHEAP,HEAP_GENERATE_EXCEPTIONS+HEAP_ZERO_MEMORY,rax
mov rdx,rax
add rax,_MEMORY_ALIGN
and rax,-_MEMORY_ALIGN
mov [rax-size_t],rdx
ret

_free:
mov rax,[rax-size_t]
invoke HeapFree,_HHEAP,HEAP_GENERATE_EXCEPTIONS+HEAP_ZERO_MEMORY,rax
ret




[dedndave]

sorry, drizz, i don't speak "rax"   :lol

[hutch--]

Rob,

To address your original question, there is no overriding reason to make a buffer divisible by 4. As Erol mentioned allocated memory is already 8 byte aligned and the solution is to simply make your buffer large enough to fit the data you want to put into it. Memory is rarely ever allocated in very small amounts at an OS level so there is no reason why you should have to trim its size down to an arbitrary byte count. Now this means that if you need 15k there is no loss at allocating 16k.

If you need to have memory aligned by greater than 8 bytes, add the extra onto the buffer length then align a location near the beginning of the start address and pass back a pointer to that location. When you deallocate the memory later make sure you deallocate the original memory pointer, not the aligned location. This matters if you want to use later SSE opcodes where they require 16 byte alignment.

[cmpxchg]

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  dec   eax              ; make sure # is either aligned to lower value or not aligned
  and   eax, -2          ; align down, has no effect if aligned to lower
  add   eax, 2           ; += 2


  number 0, 2byte aligned will result in 0
  number 1, 2byte aligned will result in 2
  number 2, 2byte aligned will result in 2
  number 3, 2byte aligned will result in 4
  number 4 2byte aligned will result in 4
                                                   

[dedndave]

why not...

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        inc     eax
        and     eax,-2
INC/DEC only works for 2-alignment
larger than that, and you are back to...

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        add     eax,M-1
        and     eax,-M