Or ax 1234h or 0ffh

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WebMOV AX,[BX+DI+1234H] 100010 D W MOD REG R/M Displacement Opcode: 100010 D: 1 Must be 1, dest AX specified by REG W: 1 16 bit transfer MOD: 10 16-bit displacement REG: 000 AX R/M: 001 [BX+DI+DISP] Machine instruction is: 1000 1011 1000 0001 0011 0100 0001 0010 8 B 8 1 3 4 1 2. WebTranscribed Image Text: Q2) if DS-3456H,SS-234AH, IP ADFCH Mov [2345H], 5675H Mov AX, 1234H ADD AX, [2345H] The final value of AX is A) 3579H b) 68A9H C) 3579D d) 1234H …

微型计算机技术与应用第3章 Pentium的指令系统 P123.ppt-原创力 …

WebNo, AX cannot store the value +32,768 because it is larger than a signed word. To determine if a value is valid or not, after the negation you would need to check the status flag to see … Webif using BX, the operand is defaultly located in Data segment,then: PA= (DS)*10H+ (BX)+ (DI)/ (SI)+disp if using BP, the operand is defaultly located in stack segment,then: PA= (SS)*10H+ (BP)+ (DI)/ (SI)+disp 16 f Based Indexed Relative Addressing MOV [BX+DI+1234H], AH 88 ;assume (ds)=4000h, (bx)=0200h, (di)=0010h ;machine code:88A13412h A1 how to rewrite cos x in terms of y

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WebDec 31, 2024 · This instruction does more or less the opposite of MOV AX, [BX]: It writes data to memory. This time, the data in the register BX is written to the memory specified by the registers DS and AX. Because on x86 systems the low byte of some data is stored in the first byte and the high byte is stored in the second byte, the value 1234h would be ... WebFeb 23, 2024 · Types of addressing modes: Register mode – In this type of addressing mode both the operands are registers. Example: MOV AX, BX XOR AX, DX ADD AL, BL. … WebExample: MOV AL, 0FFh DAS RET ; AL = 0FFh (-1) ; AL = 99h, CF = 1. CZSOPA rrrrrr. Decrement. Algorithm: operand = operand - 1 Example: ... Example: MOV AX, 1234h PUSH AX POP DX ; DX = 1234h RET 8086 instructions. Page 41 of 53. CZSOPA unchanged. Push all general purpose registers AX, CX, DX, BX, SP, BP, SI, DI in the stack. Original value of SP … northern arc burgess hill development

微机原理与接口技术期中考试题(带答案)_百度题库

Webanswer bank (DX=FFFFh, AX=80C0h / shl ebx, 3 / shl ebx, 5) 18. What will be the hexadecimal values of DX and AX after the following instructions have executed? mov ax, 3456h mov … WebMOV AX,BҮТE1 AX=1234H AX= d. ADD DL,BYTE DX=0122H DX= e. INC DX DX=FF12H DX= f. INC DL DX=FF23H DX= g. XCHG AL,AH AX=1234H AX= h. SUB CL,CH CX=3451H CX= Question For each of the following unrelated instructions, show the result in the destination operand. Indicate if the instruction is invalid. Assume BYTE1 is defined as DB 05. how to rewrite a songWeb微机系统与汇编语言复习微机系统与汇编语言简答题1. 设a186,b273q,c0bbh,它们之间的关系是 a.abc b.abcc.abc d.abc2. 80868088的存贮器组织是将存储器划分为段,可作为段的起始地址是 a.185a2 northern arc capital wikipedia

"WebEngineering Computer Science 2. Choose the assembler codes to move the value 5678h from physical memory location: 1234h to accumulator (DS=100h). a) mov ax, 1234h b) mov ax, 5678hc) mov ax, (1234h) d) mov ax, [5678h]e) mov ax, [234h] D mov ax, [4678h] 2. Choose the assembler codes to move the value 5678h from physical memory location: … " - Or ax 1234h or 0ffh

Or ax 1234h or 0ffh

Explain the instruction set of 8086 with examples? - Answers

WebJZ JNC JP JA JGE JLE Hand assemble the following codes: MOV BX, AX MOV BX, 01234H MOV BX [BP] MOV [SI+5] BX MOV [BX+SI], AL MOV AH, XYZ; Assume that XYZ is located … WebORG 100h LEA AX, m RET m DW 1234h END AX is set to: 0104h. LEA instruction takes 3 bytes, RET takes 1 byte, we start at 100h, so the address of 'm' is 104h. LES REG, memory Load memory double word into word register and ES. Algorithm: REG = first word; ES = second word; Example: ORG 100h LES AX, m RET m DW 1234h DW 5678h END AX is set …

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http://www.sce.carleton.ca/courses/sysc-3601/s14/SYSC3601-Slides-03-Intel%20Addressing%20Modes%20and%20Instruction%20Encoding.pdf WebEngineering Computer Engineering نقطة واحدة let AX=1235H,SS=1233H and SP=1234H and after the execution POP AX the SP is 1234H O 1238H 1236H 1232H O نقطة واحدة after …

Web微机原理与接口技术期中考试题(带答案)_试卷_中考 Weba. mov ax，12h b. mov ax，[1234h] c. mov ds：[1234h]，12h d. mov al，ds：[1234h] 5. 下面源程序语句中，_____包含有寄存器寻址方式。 ... and bx，0ffh and bh，[bx] 上述指令序列运行后，bx的内容是（）( b ) a. 0305h ( bcd ) a. mov al，00 b. and al，00 c. xor al，al d. sub al，al 10. 下面可以形成 ...

WebNov 11, 2024 · 指令执行前: (SS)=5000H, (BP)=3000H, COUNT=2040H, (AX)=1234H 指令执行后：EA=5040H PA=55040H (55040H)=5548H (AX)=5548H MOV AX, [BX][SI] 或 MOV AX, [BX+SI] 执行指令前： (DS)=3200H, (BX)=0456H, (SI) =1094H (334EAH)=4567H 执行指令后： EA=14EAH PA=334EAH (AX)=4567H 基址加变址寻址方式 MOV AX,[BX+SI] 第3章 … Web.data myByte BYTE 0FFh, 0 .code mov al,myByte ; AL = mov ah, [myByte+1] ; AH = AL = FFh AH = 00h .data var1 DWORD 10000h var2 DWORD 20000h .code ; ---EAX--- mov eax,var1 ; a.) add eax,var2 ; b.) add ax,0FFFFh ; c.) add eax,1 ; d.) sub ax,1 ; e.) a.) 00010000h b.) 00030000h c.) 0003FFFFh d.) 00040000h e.) 0004FFFFh Sets with similar terms

WebMOV AX, 1234h RET ; return to caller. p1 ENDP CZSOPA unchanged CBW No operands Convert byte into word. Algorithm: if high bit of AL = 1 then: zAH = 255 (0FFh) else zAH = 0 Example: MOV AX, 0 ; AH = 0, AL = 0 MOV AL, -5 ; AX = 000FBh (251) CBW ; AX = 0FFFBh (-5) RET CZSOPA unchanged Clear Carry flag. Algorithm: CF = 0 8086 instructions Page 6 of ...

WebAs far as the CPU is concerned, or eax, -1 depends on the previous value of the eax register, which lengthens the code's dependency chain and will significantly decrease performance … northern architectural productshttp://site.iugaza.edu.ps/ayash/files/Assembly-Chapter7-Part2.pdf northernarc.com northern arches food services inc spokane waWeb摘要 ds=091dh 孙德文微型计算机技术答案 ds=091dh 已知ds=091dh,ss=1e4ah,ax=1234h,bx=0024h,cx=5678h,bp=0024h,si 微型计算机技术第四版答案孙德文 northern archers of sydneyWebanswer bank (DX=FFFFh, AX=80C0h / shl ebx, 3 / shl ebx, 5) 18. What will be the hexadecimal values of DX and AX after the following instructions have executed? mov ax, 3456h mov dx, 12h mov bx, 10h div bx Choose one answer. a. DX=0006h, AX=2345h b. DX=1234h, AX=0056h c. DX=0056h, AX=1234h d. cannot be determined 19. how to rewrite a pdfWebmov AL,0FFH mov AX,0 inc AL dec AX ∗Related instructions jc jump if carry (jump if CF = 1) jnc jump if no carry (jump if CF = 0) ∗Carry flag can be manipulated directly using stc set … northern archaeological consultancyWeb微机原理和接口技术习题答案解析第1章微机运算基础习题和思考题1. 请完成以下计算:17466d10101110.10101bae. a8h10101110101.01011b 1397.344d575.58h4bch010b bcd2. how to rewrite expressions into radical form