On Friday 28 June 2002 06:23 pm, Flower wrote: > On Fri, 2002-06-28 at 15:22, fila wrote: > > 0x80004a8 <main+24>: addl $0xc,%esp > > 0x80004ab <main+27>: movl $0x1,0xfffffffc(%ebp) > > 0x80004b2 <main+34>: movl 0xfffffffc(%ebp),%eax > > [...] > > > We can see that when calling function() the RET will be 0x8004a8, > > and we want to jump past the assignment at 0x80004ab. The next > > instruction we want to execute is the at 0x8004b2. A little math > > tells us the distance is 8 bytes. > > > > si abia aici am incheiat citatul. sa-mi explice si mie cineva cum > > ii da lui 8 bytes ca io cred ca am lipsit la lectia asta in scoala > > primara. > > Pentru ca el/ei/ea socoteste doar instructiunea peste care _sare_ > adica de la 0x80004aa pana la 0x80004b2. > b2-aa=8
ar fi fost misto asa dar ce te faci ca nu e vorba de aa ci de ab si atunci : b2 - ab = 7 Fila --- Pentru dezabonare, trimiteti mail la [EMAIL PROTECTED] cu subiectul 'unsubscribe rlug'. REGULI, arhive si alte informatii: http://www.lug.ro/mlist/
