glc bx u

=: Month: Day al=00 success=FF invalid2 c Take time ch:cl=: minutesDh:dl= seconds: 1/100 secondsSet time ch:cl=: MinDh:dl= seconds: 1/100 seconds al=00 Success=FF invalid2E disk Automatic read-write flag al=00 off flagAL=01 Open Flag2F fetch the first address of the disk buffer es:bx= buffer first address30 Take DOS version number ah= release number, al= version31 end and reside al= return codedx= Dwell Area Size33Ctrl-break detectional=00 Take status=01 status (DL)Dl=00 off detection=01 Open de

complexity of O (N) (it only refers to the complexity of merging intervals, and the complexity of sorting previously facing intervals is O (nlogn )). # Include # Include String . H># Include # Include # Define Zero 1e-8 # Define Maxd 510 Int N, R [maxd]; Double BX, by, CX [maxd], CY [maxd], Cr [maxd], LX [maxd], RX [maxd]; Int CMP ( Const Void * _ P, Const Void * _ Q){ Int * P = ( Int *) _ P; Int * Q = ( Int *) _ Q; Return Lx [* p] 1 : 1 ;} D

I want to write a fully functional character driver, but I still have to consider it. Record your initial attempt to drive the character In the absence of an operating system, you need to do the following to make the system respond to the keyboard: 1. Writing a keyboard interrupt program: When you press the key, the program will enter 2. Initialization of 8259a: When you press the key, the 8259a hardware is interrupted, allowing the CPU to enter the keyboard interrupt program 3. In the interrupt

Set (ax) = 5, (BX) = 3, (CX) = 2, run the following program section, (ax) = lop: ror BX, cl adc ax, BX loop lop how does ror shift? Thank you. Ror is the circular right shift command, which shifts the entire right of the destination operand from the number of digits specified by the source operand. The removed bits are refilled to the left blank bits and moved

jump have unconditional jump: JMP. and conditional jump, JX. There are 19 instructions on X. DATA SEGMENT XX db 5 YY db? DATA ENDS Stack SEGMENT stack DB DUP (?) STACK ENDS CODE SEGMENT assume Cs:code, Ds:data, Ss:stack START: mov ax, DATA mov DS, ax ; Core code segment MOV al, XX CMP al, 0; X-0 to establish the Jge Biger of sign position; X>=0 jump to Biger MOV AL,-1; X Enter 5 and return 1. No problem 、、、 Eg: programmatic implementation: A1+....+A10 o

the NUL device driver? What dos can find indicates that it must have saved this address, and all the places where it is saved will be OK. To get this address, we have to use a non-public dos call with 52 h function. Here we only need to provide a description of what we need. If the reader needs a complete description of this function, you can contact me. Int 21 H function 52 h:Entry: Ah = 52 h Exit: Es: BX pointing to important data List table In the

============================================Function: determines whether string 1 is a substring of string 2.; Entry:; Cx = length of string 1; BX = string 2 Length; SI = start address of string 1; DI = start address of string 2Export:; BP> = 0 string 1 is the substring of string 2, and BP = the position where string 1 first appears in string 2 (starting from 0); BP =-1 string 1 is not a substring of string 2; BP =-2 string 1 is longer than string 2;

Chapter 4 first programExperiment 3 programming, compilation, connection, and tracking1. Save the following program as a t1.asm file and generate the executable file t1.exe.Assume Cs: codesgCodesg segmentMoV ax, 2000 hMoV SS, axMoV sp, 0Add SP, 10Pop axPop BXPUSH AXPush BXPop axPop BX MoV ax, 4c00hInt 21 h Codesg ends End 2. Use debugto track the execution process of t1.exe and write the internal Content on the content and stack topDebug t1.exe-RAx =

_ Processor detection schemes _ By Richard C. leinecker [Listing one] ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ; Detect the processor type -- by Richard C. leinecker ;; ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; _ Ptext segment para public 'code' Assume Cs: _ ptext, DS: _ ptext Public _ Processor ; This routine returns the processor type as an integer value. ; C prototype ; Int processor (void ); ; Returns: 0 = 8088, 1 = 8086,

function as the C language? ; Stack segmentStsg segment Stack's'DW 64 DUP (?)Stsg ends ; Data SegmentData SegmentRlength equ 2048Fname dB 'access. log _ ', 0Hit DB 'Hit $'Fd dw? ; File handleResault dB 'count: $ '; Result promptCount dd 0; Save the resultDisflag db 0; display flagBuff dB 5000 DUP (0); buffer zoneData ends ; Code segmentCode segmentMain proc farAssume Cs: code, DS: data, SS: stsg, ES: NothingMoV ax, DataMoV ds, ax; My code starts:MoV ah, 3DH; open the fileLea dx, fnameMoV Al, 00

Chapter 1 registers (CPU operating principles) 2.1 General registersAll registers of the 8086cpu are 16-bit and can be stored in two bytes. The four registers ax, BX, CX, and dx are usually used to store general data and are called General registers.A 16-bit register can store a 16-bit data.What is the maximum value of data stored in a 16-bit register?The registers in the previous generation of 8086cpu are all 8-bit. To ensure compatibility, the progr

After entering debug, first use the command R to view the current DS content and record this value, which is assumed to be X. Compile and Debug programs that complete the following functions under debug. 1): Transfer 30 h words from f000: 2000h to memory X: 1000h Here we copy the data of a unit to another unit, which is continuous data, so we must use a loop. Because of the differences between the two Ds contents, when we copy the data again, to copy ds, see the following code. 1 MOV CX,3

based on the memory size. Let's take a look at the example and change the first four characters of each of the following words (display, brow, replace, modify) to uppercase. Assume Cs: codesg, DS: datasg, SS: stacksg; Stacksg segment DW 0, 0, 0, 0, 0, 0, 0; Stacksg ends Datasg segment DB '1. display .......' DB '2. brow ..........' DB '3. Replace .......' DB '4. Modify ........' Datasg ends Codesg segment Start: mov ax, stacksg MoV SS, ax MoV sp, 16; Link stacksg to SS (define stack and initia

Getting started with crazy summer vacation Study Notes (1) Reference: Chapter 1 and Chapter 2 of Assembly Language I. Basic Knowledge 1. Working Process of programming in assembly language The essence of assembly is to make it easier for programmers to use assembly commands to represent binary code. Assembly instruction ----------------> compiler ----------------> machine code ----------------> Computer execution MoV ax, BX 100010001 ........

flag, 0 to-use DOS, 1 to-use BIOS example. MODEL TinyUse_bios_flag EQU 1;-----------Stack Area---------------. STACK 500;--------------------------------------. CODE;*************　@@@@@@@@@@@@@@　***************Start proc nearMOV bx,80h index command line datamov al,[bx]; get size of string variableMOV Ax,csmov ds,ax; reset data segmentMOV psp_seg,es; save PSP Addressmov es,ax; reset extra SegmentCMP al,4;

1. Target positioning 1.1 Introduction to classification, positioning and testing -Image classificationImage classification, is to give you a picture, you determine the target category, such as cars, cats and so on.-Classification with localizationPositioning classification, not only to determine the target category, but also to output the position of the target object, such as the box up.-DetectionDetection, there may be multiple objects in the picture, you need to find them out. 1.2 Position

hahahaBFS+SPFA.#include #include#include#include#include#defineMAXN 35#defineMAXV 100500#defineMaxe 400500#defineINF 100000007using namespacestd;structedge{intV,W,NXT;} E[maxe];intn,m,qq,map[maxn][maxn],dp[maxn][maxn][4],step[maxn][maxn][4][4],tot=0, nume=0, g[maxv],v[maxn][maxn][4];intdt[maxn][maxn],dx[]={-1,1,0,0},dy[]={0,0,-1,1},DIS[MAXV];intex,ey,sx,sy,tx,ty,s,t;BOOLVis[maxv];queueint>Q;voidAddedge (intUintVintW) {e[++nume].v=v; E[NUME].W=W; E[NUME].NXT=G[u]; G[u]=Nume;}voidReset1 () { for(i

# coding:utf-8__author__ = ' similarface ' #datalink =http://www2.informatik.uni-freiburg.de/~cziegler/bx/ Bx-csv-dump.zip "bx-users[" User-id ";" Location ";" Age "]bx-books[" ISBN ";" Book-title ";" Book-author ";" Year-of-publication ";" Publisher ";" Image-url-s ";" Image-url-m ";" Image-url-l "]

5. The executable file consists of descriptive information and programs, which are derived from the assembly instructions and the defined data in the source program, while the descriptive information is mainly compiled and connected by the program to process the pseudo-instructions in the source program. Where the CPU starts executing the program, it can be specified in the source program as "End label". 6. Register: REG contains: Ax,bx,cx,dx,ah,

For example: When we search for the footfall home in Baidu, the parameter behind the URL is Http://www.baidu.com? CL = 3 WD = % BD % C5 % B1 % be % D6 % AE % BC % D2 To restore % BD % C5 % B1 % be % D6 % AE % BC % D2 to Chinese, use the following function.Response. Write urldecode ("http://www.baidu.com/s? CL = 3 WD = % BD % C5 % B1 % be % D6 % AE % BC % D2 ")'================================================ =========='Function name: urldecode'Usage: URL Decoding'==========================