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Article #2189Re: Równie prosta zagadka
From: Bernard
Date: Fri, 09 Apr 2010 12:47
Date: Fri, 09 Apr 2010 12:47
25 lines
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1491 bytes
On 2010-04-09 10:52, Bernard wrote: > Klasycznie robi się to np. tak: > > ; EDI - adres poczatku, ECX - l. bajtow > > lea esi, [edi + ecx - 1] > rlop: > mov al, [edi] > xchg al, [esi] > stosb > dec esi > cmp esi, edi > jb rlop Popełem błąd w warunku wyjścia z pętli - powinno być ja. Ładniejsze jest chyba jednak coś takiego - tylko goła pętla, 5 instrukcji, 3 rejestry: xlop: lodsb xchg al, [esi + ecx - 2] mov [esi - 1], al sub ecx, 2 ja xlop
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