Assembly language--[BX] and loop instructions

[BX]

Because in the compiler mov ax,[0] instruction will be considered by the compiler to be MOV ax,0 so introduced the concept of [BX]

mov [bx],ax means to assign the value of the (DS) *16+ (BX) memory location to Microsoft Dynamics AX

Use the same as before Ds[address]

Small Tips

1 ;How to use [BX]2 movax,1000h3 4 movbx,2000h5 6 movDs,ax;assign a value of 2000H to the DS Data segment register7 8 mov[Bx],ax;assign the value of the AX register to the memory unit 21000H location9 Ten ;Other ways One movAxds:[0];assign the value of the memory Unit 20000H location to Microsoft Dynamics AX

Loop command (Loop)

The format of the directive is: loop designator, when the CPU executes the Loop command, two steps are taken:
① (CX) = (CX)-1;
The ② determines the value in CX, does not zero, and then goes to the label to execute the program, and if zero, executes down.

Usually we use loop instruction to realize the loop function, thenumber of cycles stored in CX.

Example: Using a program to calculate 211

1AssumeCS:Code2 Code Segment3 Start:movAx24       movCx One      ;set the number of cycles to one5       S:  AddAx,ax;each time the operation is performed, the Cx-16Loop s;verifies that the value of CX is 0, executes the following program for 0, or jumps back s (s can be any string) to execute7       movax,4c00h;Program returns8       int21h9Code ends;end of code snippetTenEnd Start

Example 2: Calculate 123 x236 with addition, the result exists in AX.

1AssumeCS:Xinge;Locate the code snippet,2Xinge segment;The code snippet can be any name3 Start:movAx04       movCx1235       Jiaxin:6         AddAx236   ;operations to Loop on7 Loop Jiaxin8       movax,4c00h9       int21hTen code ends OneEnd Xinge

Combined application of loop and [BX]

Thinking: Computes the ffff:0~ffff:b of the data in the cell and stores the results in DX.

First look at a wrong: The cause of the error, only defined dl++, overflow does not carry

1AssumeCS:Code2 Code Segment3 Start:    movAx,0ffffh;The compiler does not support the beginning of a letter4     movDs,ax5     movBx06     movDl07     movDh08     movCx A9     Ten     S:AddDL,[BX] One       IncBX A Loop S -        -       movax,4c00h the       int  + - code ends -End Start

The right

1AssumeCS:Code2 Code Segment3 Start:    movAx,0ffffh;The compiler does not support the beginning of a letter4     movDs,ax5     movBx06     movDx07     movCx A8     9     S:movAL,[BX];assign value to lowTen       movAh0     One       AddDx,ax;add with 16 bits so that it does not overflow A       IncBX - Loop S -        the       movax,4c00h -       int  + - code ends -End Start

Segment Prefix

The "DS:", "CS:", "SS:", or "es:" that appear in the instruction that accesses the memory unit to explicitly indicate the segment address of the memory unit, called the segment prefix in assembly language.

Use of segment prefixes:

Example: Copy data from a memory ffff:0~ffff:b segment into a 0:200~0:20b cell.

Method One: Continuously modify the DS value and then assign the value

Method Two: Use the segment prefix es: (es is the extra segment register), so that you can eliminate the constant change in the DS value

1AssumeCS:Code2 Code Segment3     movcx,40h;cycle times, 3f+14     movAx05     movDs,ax6     movbx,200h7     movDx08     S:9     mov[Bx],dxTen     IncDX One Loop S A      -     movax,4c00h -     int21H the code ends -End

answer to the first question

1AssumeCS:Code2 Code Segment3     movcx,40h4     movAx -    ;another address changed to5     movDs,ax6     movBx0    ;BX can start from 0, the same as to push the value of the memory unit7     S:8     mov ds:[Bx],bx9     IncBXTen Loop S One      A     movax,4c00h -     int21H - code ends theEnd

answer to the second question

Assembly language--[BX] and loop instructions