division in assembly language programming

From how to establish the source file to the debugging for a brief introduction, and with examples. Because of my limited level, please point out in the following areas where there are errors and can be further modified assembly language on the machine process:First, the pre-machine software preparation:MS-DOS operating system (e.g. MSDOS6.22, MSDOS7.0, etc.)Text editors such as: Edit.com, TURBO. EXE, TC.

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

the interrupt process. In 8086CPU, for example, after receiving the interrupt information, the action to be taken:1. (from the interrupt message) find the interrupt bytecode2. Value of the flag register into the stack3. Set the 8th TF and 9th bit if value of the flag register to 04.CS value into the stack5.IP value into the stack6. From the memory address for terminal type code, and interrupt type code *4+2 two characters read the interrupt handler entry address, set CS IPInterrupt handlersSinc

assume CS:CODESG, es:table, ds:datadata segment DB'1975','1976','1977','1978','1979','1980','1981','1982','1983'DB'1984','1985','1986','1987','1988','1989','1990','1991','1992'DB'1993','1994','1995'; 21 year DD -, A,382,1356,2390,8000,16000,24486,50065,97479,140417,197514DD345980,590827,803530,1183000,1843000,2759000,3753000,4649000,5937000; 21 Divisor DW3,7,9, -, -, -, the, -,476,778,1001,1442,2258,2793,4037,5635,8226DW11542,14430,15257,17800; 21 Divisor data endstable Segment DB +DUP ('Year su