mips assembly language

Machine language is a set of machine instructions, machine instructions is a column of two-level digital, the computer will be transformed a row of high and low voltage, so that the computer's electronic devices are driven, to operate, each CPU has its own machine instruction set, which is the rule, also called machine language, the early programmers write code, are holes, Programmers will use 0, 1 digital

Writing assembly language with DOSBox under WIN8 DOSBox is a DOS simulation program that can be easily ported to other platforms , Therefore, it can be used to write assembly language in the WIN8 system . the relevant download links are provided below:DOSBox Download: Http://pan.baidu.com/s/1qWwkss0

ObjectiveBefore starting this series of articles, I would like to say, for the antique-grade 8086 compilation in the end, there is no need to learn. Here I say I want to start from 8086, rather than from the Win32 compilation or WIN64 assembly to start learning directly.The more advanced the development technology is, the deeper it will be for the underlying package. I learned the purpose of assembly

4.0 using subroutines with interrupts have mastered the assembly language? Yes, you can now decipher the secret in someone else's code. However, one of the important things we did not mention was that the program and the interruption. These two things are so important that your program is almost impossible to leave them. 4.1 Sub Program In high-level languages we often use subroutines. In advanced langua

+di+idata]The following instruction is incorrect:mov Ax,[bx+bp]mov Ax,[si+di]3. Just again ' [...] ' With register BP addressing, the segment register defaults to SS.For example, the following command:mov ax,[bp] (AX) = ((ss) *16+ (BP)) mov Ax,[bp+idata] (AX) = ((ss) *16+ (BP) +idata) mov Ax,[bp+si] (AX) = ((ss) * 16+ (BP) + (SI)) mov ax,[bp+si+idata] (AX) = ((ss) *16+ (BP) + (SI) +idata)8.2 Where is the machine instruction processing data?when executing machine instruc

instructions:Format:MOVSB; MOV string byteFunction: 1) ((es) *16+ (di)) = ((ds) *16+ (SI))ie: mov es:[di],byte ptr ds:[si]2) when df=0, (si) = (si) +1(di) = (DI) +1That is: Inc siInc diWhen Df=1, (si) = (si)-1(di) = (DI)-1namely: Dec siDec diFormat:MOVSW; MOV string wordFunction: 1) ((es) *16+ (di)) = ((ds) *16+ (SI))ie: mov es:[di],word ptr ds:[si]2) when df=0, (si) = (si) +2(di) = (DI) +2That is: Add si,2Add di,2When Df=1, (si) = (si)-2(di) = (DI)-2That is: Sub SiSub diGeneral MOVSB/MOVSW in

; Function "show hellow word". 586. Model flat, stdcall; model, inter-function Communication;----------------------------------------------Option Casemap: none;Include windows. incInclude kernel32.incInclude user32.inc; header file;-------------------------------------- ;-------------------------------------Includelib kernel32.libIncludelib user32.lib; Library file call;--------------------------------------. Data; data segmentStr1 DB "Hello word! ", 0Str2 DB "mynote", 0. Data?Str3 DB ?, 0; no

: address provided for Stack operations. 4. I/O control logic: function: input/output operations. 5. CPU program execution process:1. Run Command 1, Run Command 2, Run Command 2, and run command 3 .... Low CPU EfficiencyTo overcome these disadvantages:8086 of commands are executed in parallel. The command queue is used.--------- Biu Bus Interface Unit EU Execution UnitBiu is responsible for reading the command to the queue, and waits until it knows it is full. EU is responsible for pulling the c

(i) the foregoingYou can modify the IP, or both CS and IP instructions are collectively referred to as transfer instructions .The transfer behavior has the following categories: When you modify IP only, it is called intra-segment transfer, for example: JMP ax. Simultaneous modification of CS and IPs is called inter-segment transfer, for example: jmp 1000:0. because the transfer instruction has different scope for IP modification , the transfer is divided into: short transfer an

1. Title: Uppercase and lowercase alphabetic characters interchange2. Requirements: Enter a character from the keyboard, if the character is a carriage return, exit the program directly, if it is a lowercase letter, it is converted to uppercase and displayed, if it is uppercase, is converted to lowercase letters and display, if it is non-alphabetic characters, display a prompt message, and wait for the user to reenter characters.3. Tip: First determine whether it is a carriage return, if not, wh

From the "Assembly Language" forumWhat is the meaning of the argument here, do an experimental test on it, the largest 64K should be no one against it is the smallest, some say 0 byte, some say 16Bytes, some say other, that is how much, we see this programAssume Cs:code, Ds:dataData segmentData endsCode segmentStartMOV ax, dataMOV ds, axMOV bx, CSSub ax, BX code endsEnd StartHere, I first define an empty da

Wang Shuang assembly language Detection Point 10.5 (1):After the following program executes, what is the number in AX?Assume Cs:codeStack segmentDW 8 DUP (0)Stack endsCode segmentStartMOV Ax,stackMOV Ss,axMOV sp,16MOV Ds,axMOV ax,0Call word ptr ds:[0eh];Inc AXInc AXInc AXMOV ax,4c00hint 21hCode endsEnd StartAnswer: ax=3Analysis:1, the subject stack segment with DW 8 DUP (0) defines 16 bytes of memory 0 of t

Compilation implements a small-to-large bubble sortThe main function is implemented by C, and the sort function is written in assembly language.#include int buffer[256]; Data buffersint bufferlen=0; Data buffer Data countextern sort (int *buffer,int bufferlen); Assembler function Sort InterfaceInput data from the keyboard ********int inputdatafromkeyboard (int *pbuffer){int ncnt=0;int x;printf ("Data entry

Problem Introduction:Linux elf files are very difficult to understand at first, some people may go to see "linkers and Loaders", this book is really good, but there is no detailed explanation of many details, especially from the assembly language perspective. I read this book a lot of places still do not understand, and then I read the IBM360 computer Assembler design document, which detailed the two times

BX, si, Di, bp1, in 8086CPU, only these 4 registers (BX, BP, Si, di) can be used in the "[...]" to address the memory unit.2. In "[...]", these 4 registers (BX, BP, Si, di) can appear either individually or only in four combinations:bx and Si, bx and di, BP and Si, BP and di1 the correct instructions:2 movAX,[BX]3 movAx,[si]4 movAx,[di]5 movAX,[BP]6 movAx,[bx+si]7 movAx,[bx+di]8 movAx,[bp+si]9 movAx,[bp+di]Ten movAx,[bx+si+idata] One movAx,[bx+di+idata] A movAx,[bp+si+idata] - movAx,[bp+di+idata