addressing-mode

What is the meaning of x86 instruction “call dword ptr ds:[00923030h]”?

与世无争的帅哥 提交于 2019-12-03 03:35:18
What does the following x86 assembler instruction do? call dword ptr ds:[00923030h] It's an indirect call I suspect, but exactly how does it compute the address to the call? [EDIT] Updated Whenever you see a memory operand that looks something like ds:0x00923030 , that's a segment-relative addressing mode. The actual address being referred tp is at linear address 0x00923030 relative to the base address of the ds segment register. Memory segmentation in the x86 architecture is somewhat confusing, and I think Wikipedia does a good job of explaining it. Basically, x86 has a number of special

The different addressing modes of CUDA textures

耗尽温柔 提交于 2019-12-03 02:51:14
I am using a CUDA texture in border addressing mode ( cudaAddressModeBorder ). I am reading texture coordinates using tex2D<float>() . When the texture coordinates fall outside the texture, tex2D<float>() returns 0 . How can I change this returned border value from 0 to something else? I could check the texture coordinate manually and set the border value myself. I was wondering if there was CUDA API where I can set such a border value. As mentioned by sgarizvi, CUDA supports only four, non-customizable address modes, namely, clamp , border , wrap and mirror , which are described in Section 3

What addressing mode is used in 'mov cx, [bp+6]'?

霸气de小男生 提交于 2019-12-02 08:32:00
问题 What addressing mode is used in "mov cx, [bp+6]"? The processor is intel 8086. I am studying "Microprocessor and Interfacing" by Douglas V. Hall. I know its memory addressing mode. But not sure whether its based addressing mode or index addressing mode? 回答1: [bp+6] is the based addressing mode. From the original 8086 docs: In based addressing, the effective address is the sum of a displacement value and the content of register BX or register BP. Indexed addressing mode is similar but with the

What addressing mode is used in 'mov cx, [bp+6]'?

梦想的初衷 提交于 2019-12-02 04:11:05
What addressing mode is used in "mov cx, [bp+6]"? The processor is intel 8086. I am studying "Microprocessor and Interfacing" by Douglas V. Hall. I know its memory addressing mode. But not sure whether its based addressing mode or index addressing mode? [bp+6] is the based addressing mode. From the original 8086 docs: In based addressing, the effective address is the sum of a displacement value and the content of register BX or register BP. Indexed addressing mode is similar but with the SI or DI registers. Basically, you have the following modes: Direct memory accessing like [1234] . Register

Do terms like direct/indirect addressing mode actual exists in the Intel x86 manuals

流过昼夜 提交于 2019-12-02 00:42:44
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, Indirect, Indexed, Based , and some more. But the blogs use inconsistent names when referring to these

Assembly (,%eax,4)

吃可爱长大的小学妹 提交于 2019-12-01 19:49:03
If one of my command lines says: jmp *0x804a180(,%eax,4) what does that mean? I ask specifically because there is no value before the first comma and I'm not sure exactly what the * before the address means. This instruction jumps to the location whose value is located at the address calculated as %eax * 4 + 0x804a180 . The * is used in AT&T syntax to indicate indirect jumps and calls. It basically means "jump to the location pointed to by this, instead of the value of this". It is useful to differentiate the following instructions: jmp myAddress # Jumps to the location myAddress jmp

How to load address of function or label into register in GNU Assembler

前提是你 提交于 2019-12-01 19:01:06
I am trying to load the address of 'main' into a register (R10) in the GNU Assembler. I am unable to. Here I what I have and the error message I receive. main: lea main, %r10 I also tried the following syntax (this time using mov) main: movq $main, %r10 With both of the above I get the following error: /usr/bin/ld: /tmp/ccxZ8pWr.o: relocation R_X86_64_32S against symbol `main' can not be used when making a shared object; recompile with -fPIC /usr/bin/ld: final link failed: Nonrepresentable section on output collect2: error: ld returned 1 exit status Compiling with -fPIC does not resolve the

Trying to understand this short assembler instruction but I don't understand

别等时光非礼了梦想. 提交于 2019-12-01 18:11:35
问题 We had a task, given was an assembler instruction of a 2-addressing machine: mov 202, 100[r1+] Note down a minimal assembler instruction sequence which replaces this instruction (see above) where n[rx+] : register indexed by post increment; n is index value and rx is register x single numeric value: directly addressed / stored The addresses we are supposed to use are: rx - register direct addressing [rx] - register indirect addressing #n - directly addressing And we are only allowed to use

Why does the mov instruction have use ax instead of two segment registers directly?

陌路散爱 提交于 2019-12-01 05:19:41
I see code like: mov ax, cs mov ds, ax mov es, ax Why can't I just compress this to: mov ds, cs mov es, cs Is the first way faster since its using the accumulator register? But that wouldn't seem intuitive since cs and ds are segment registers . Or is there some restriction that I'm unaware of? I'm using nasm by the way. You can't mov segment register to segment register -- there's no instruction for it. There is only so much room in a processor for the microcode for all its instructions. So one general instruction is often preferred over several special purpose ones for rarely uused

Why does the mov instruction have use ax instead of two segment registers directly?

廉价感情. 提交于 2019-12-01 03:06:02
问题 I see code like: mov ax, cs mov ds, ax mov es, ax Why can't I just compress this to: mov ds, cs mov es, cs Is the first way faster since its using the accumulator register? But that wouldn't seem intuitive since cs and ds are segment registers . Or is there some restriction that I'm unaware of? I'm using nasm by the way. 回答1: You can't mov segment register to segment register -- there's no instruction for it. 回答2: There is only so much room in a processor for the microcode for all its