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Is it because memory addressing inside lea is free? Edit 1:also, how does[rax + rax*8]get translated into machine code? Does it gets compiled down to additional 2 instructions (shl, rbx, rax, 3; add rax, rax, rbx;), or something else? Edit 2:Surprising results...
Bevacizumab plus paclitaxel/carboplatin is predicted to become the clinical gold-standard for advanced ovarian cancer in 2011, according to a report* by Decision Resources. The addition of bevacizumab [ Avastin ; Roche, Genentech, Chugai Pharmaceutical] to the standard regimen of paclitaxel [Taxus; ...
82.2k4949 gold badges284284 silver badges500500 bronze badges Add a comment 1 Answer Sorted by: 9 I'm asking about meeting the requirements of the x64 ABI. Is it safe to blindly adjust the stack by growing it 8 bytes for a 16-byte alignment after every call. Yes, that's the w...
If equal, set ZF and load RCX:RBX into m128. Else, clear ZF and load m128 into RDX:RAX. First I replace all 5 bytes of the instruction with a jump to code cave with my emulation procedure, luckily the jump takes up exactly 5 bytes! The jump is actually a call instruction e8, but...
One of the arg-register orders he evaluated was RAX, RDX, RCX, RBX, RSI, RDI, but he found that less good than other options. (See the mailing list message linked above). It's fairly common for RISC calling conventions to pass the first arg in the first return-value reg...
push rbx mov ebx, edi# save function arg in a call-preserved regmov edi, OFFSET FLAT:guard variableforfoo(int)::x# addresscall __cxa_guard_acquire #guard_acquire(&guard_x) presumably a normal mutexorspinlock test eax, eax jne .L14#if(we won the race to do the init work) goto...