addressing-mode

问题 I read Intel manual about address calculation in 64 bit mode. Suppose we have 64 bit mode, default address size is 64 bit. Suppose also that instruction in question is preceded by address size override prefix, so address size becomes 32 bit. Now, imagine that instruction encodes memory operand with address specified by register number 0 (rax/eax/ax ...). Now the thing that I don't fully understand, whether CPU looks only at eax value and internally zero-extends it to form "native" 64 bit

问题 I have read the following in the book Programming from the Ground Up : Processors have a number of different ways of accessing data, known as addressing modes. The simplest mode is immediate mode , in which the data to access is embedded in the instruction itself. For example, if we want to initialize a register to 0, instead of giving the computer an address to read the 0 from, we would specify immediate mode, and give it the number 0. In the register addressing mode , the instruction

问题 I have read the following in the book Programming from the Ground Up : Processors have a number of different ways of accessing data, known as addressing modes. The simplest mode is immediate mode , in which the data to access is embedded in the instruction itself. For example, if we want to initialize a register to 0, instead of giving the computer an address to read the 0 from, we would specify immediate mode, and give it the number 0. In the register addressing mode , the instruction

问题 This question already has answers here : Referencing the contents of a memory location. (x86 addressing modes) (2 answers) Closed 3 years ago . I'm making a x86 asm to C code emulator.. and for my parser I am up to the bracket parsing QWORD PTR [ to ] DWORD PTR [ to ] WORD PTR [ to ] BYTE PTR [ to ] MOV X, [ to ] LEA X, [ to ] For now I will ignore.. MMWORD PTR[] XMMWORD PTR[] FWORD PTR [] TBYTE PTR [] I want to know what are all the possible mathematical arithmetic can that be placed in the

问题 I have searched for Addressing modes in IA-32,but I haven't seen any website or an article which have explained the addressing modes simply. I need an article or something which explain the matter simply by some picture of the memory during it changes and specifying the address mode by pictures. I know that in IA-32 general form of addressing follows the following form : Segment + Base + (index * scale) + displacement I want to know the exact meaning of the displacement,scale,index and

问题 I am trying to get a good grip on the LEA instruction in x86: leal (%edx, %edx, 4), %eax leal (%edx, %edx, 2), %eax Given these two lines, i know that: eax = edx + edx*4 and then eax = edx + edx*2 Two questions. First, if these instructions appear in sequence as in this example, the eax register is overwritten once the second line executes? And what exactly would be loaded into the register? Another address? Or is this doing arithmetic on the values that these registers point to? 回答1: if

问题 I am trying to get a good grip on the LEA instruction in x86: leal (%edx, %edx, 4), %eax leal (%edx, %edx, 2), %eax Given these two lines, i know that: eax = edx + edx*4 and then eax = edx + edx*2 Two questions. First, if these instructions appear in sequence as in this example, the eax register is overwritten once the second line executes? And what exactly would be loaded into the register? Another address? Or is this doing arithmetic on the values that these registers point to? 回答1: if

问题 You'll have to excuse me, I'm brand new to x86 assembly, and assembly in general. So my question is, I have something like: addl %edx,(%eax) %eax is a register which holds a pointer to some integer. Let's call it xp Does this mean that it's saying: *xp = *xp + %edx ? ( %edx is an integer) I'm just confused where addl will store the result. If %eax is a pointer to an int, then (%eax) should be the actual value of that int. So would addl store the result of %edx+(%eax) in *xp ? I would really

问题 To give a little bit of background, I wanted to study how x86 instructions are encoded/decoded manually. I came across the ModR/M and SIB bytes and it seems that understanding the x86 addressing modes is fundamental to understanding the instruction encoding scheme. Hence, I did a Google search for x86 addressing modes. Most blogs/videos that the search returned were addressing modes for the 8086 processor. Going through some of them, the different addressing modes were Register, Direct,

问题 This question already has answers here : What is the meaning of MOV (%r11,%r12,1), %edx? (2 answers) How does “mov (%ebx,%eax,4),%eax” work? [duplicate] (1 answer) Closed 2 years ago . I'm running through some code right now on gdb and I have no clue what these two instructions actually do. If anyone could help me out, I'd really appreciate it. add -0x2c(%ebp, %ebx, 4), %eax cmp %eax, -0x28(%ebp, %ebx, 4) 回答1: x86 assembly is usually much easier to understand when you write it in Intel syntax