Suppose we have a call foo statement. So when the assembler encounters a call statement it breaks it down into -
char program[3] = { 0x6a,0x0a, 0xC3 }; //push 10; ret; char (*p)(void) = program; printf(\"char = %c \\n\", p());
Let\'s say that the environment is x86. How do compilers compile the \">\" operator on 32-bit integers. Logically, I mean. Without any knowledge of Assembly.
I have a little开发者_如何学C 8086 emulator and I\'ve had a long standing bug for like 2 years now that AF does not behave properly inside of sub and add instructions.
I need help understanding endianness inside CPU registers of x86 processors. I wrote this small assembly program:
If we have: test dword ptr [eax], 2000h je label1: Is there any value other than 0 in dword p开发者_JAVA技巧tr [eax] that would make the jump take place?Instruction test works like and instruction,
This question already has answers here: 开发者_如何学Python What's the purpose of the LEA instruction?
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I remember having read a开发者_StackOverflow中文版bout it somewhere… Could anyone shed some light on this?According to Wiki, Pentium Pro. They are suprizingly old, and I wonder why do you still need
I’m currently learning about the stack (x86). I know it’s a pile of data which operates according the LIFO principle. I know the basic operations in regards to the stack are push (to add a value on
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