IBM-PC汇编语言程序设计第三章习题详解答案（国外英文资料）.doc

IBM-PC汇编语言程序设计第三章习题详解答案（国外英文资料） The IBM PC assembler language program design chapter iii answers the question Chapter 3 answers 1-20 = = = = = = = = It is known that (BX) = 637DH, (SI) = 2A9BH, displacement D = 7237H, what is the valid address in the following various addressing mode (1) there is no immediate addressing (2) direct addressing 7237H (3) there is no register addressing using the BX (4) the indirect addressing of the BX is 637DH (5) use the BX register relative to addressing (6) base address addressing 8E18H (7) the relative base address is 004FH Try to write the corresponding assembly instructions according to the following requirements. (1) ADD DX, BX (2) ADD AL, [BX] [SI] (3) ADD [BX] (4) the MOV AX, 2 a59h The ADD [h] 0524, AX (5) the ADD AL, 0 b5h Write the sixth word of the word BLOCK for BLOCK, which is sent to the DX register. The following types of addressing are required. (1) the register is indirectly addressed MOV BX, OFFSET [BLOCK], or LEA BX, BLOCK + 0AH MOV DX, [BX] (2) register relative addressing MOV SI, OAH or LEA BX, BLOCK MOV DX, [BLOCK] [BX + 0AH] (3) address addressing MOV BX, BLOCK, MOV SI, OAH MOV DX [BX] [SI] 4. (DS) = 2000 h, existing (BX) = 0100 h, (SI) = 0002 h, (20100) = 12 h, (20101) = 34 h, (20102) = 56 h, (20103) = 20103 h, ah, (21200) = 2 (20201) = 4 ch, (21202) = B7H, (21203) = 21203 h, after the test shows that the following instruction execution, AX the contents of the registers. (1) MOV AX, 1200H 1200H (2) MOV AX, BX 0100H (3) MOV AX, [1200] 4C2AH, the word unit! (4) MOV AX, [BX] 3412H, is the word unit! (5) MOV 1100 (6) MOV (7) MOV, 1100 [BX] Given (IP) = 2BC0H, (CS) = 0200H, displacement D = 5119H, (BX) = 1200H, (DS) = 212AH, (224A0) = 0600H (1) direct addressing of 7CD9H (2) indirect addressing of 1200H using the BX and register indirect addressing mode (3) the indirect addressing of 6319H is used in terms of the relative addressing method of the BX and register 6. Set the contents of the current data segment regi