User:Samupedia4

Program 1 Memory location Op-code & operand Mnemonics 0000:0100 B8 05 00 MOV AX,0005H 0000:0103 BB 02 00 MOV BX,0002H 0000:0106 01 D8 ADD AX,BX 0000:0108 CD A5 INT A5 Result after execution of above program AX = 0007H, BX = 0002H, FX = F002H, IP = 010AH CS,DS,ES,SS = 0000H D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 0 1 0 NS NZ NAC PO NC 

i.e. FL = 02H

Example 2 : If inputs are AX = 00FFH, BX = 0002H Then output AX = 0100H, BX = 0002H, FX = F016H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 1 0 1 1 0 NS NZ AC PE NC i.e. FL = 16H


Example 3 : If inputs are AX = FFFFH, BX = 0001H Then output AX = 0000H, BX = 0001H, FX = F057H, D7 D6 D5 D4 D3 D2 D1 D0 0 1 0 1 0 1 1 1 NS Z AC PE C i.e. FL = 57H


Program 2 Memory location Op-code & operand Mnemonics 0000:0100 B8 05 00 MOV AX,0005H 0000:0103 BB 02 00 MOV BX,0002H 0000:0106 29 D8 SUB AX,BX 0000:0108 CD A5 INT A5 Result after execution of above program AX = 0003H, BX = 0002H, FX = F006H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 1 1 0 NS NZ NAC PE NC 

i.e. FL = 06H


Example 2 : If inputs are AX = 0002H, BX = 0005H Then output AX = FFFDH, BX = 0005H, FX = F093H, D7 D6 D5 D4 D3 D2 D1 D0 1 0 0 1 0 0 1 1 S NZ AC PO C 

i.e. FL = 93H

Example 3 : If inputs are AX = FFFFH, BX = FFFFH Then output AX = 0000H, BX = FFFFH, FX = F046H,

D7 D6 D5 D4 D3 D2 D1 D0 0 1 0 0 0 1 1 0 NS Z NAC PE NC

i.e. FL = 46H

Program 3 Memory location Op-code & operand Mnemonics 0000:0100 B8 00 00 MOV AX,0000H 0000:0103 9F LAHF 0000:0104 CD A5 INT A5 Result after execution of above program AX = 0200H, FX = F002H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 0 1 0 NS NZ NAC PO NC 

i.e. FL = 02H

Example 2 : If inputs are AX = FFFFH, Then output AX = 02FFH, FX = F002H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 0 1 0 NS NZ NAC PO NC i.e. FL = 02H


Example 3 : If inputs are AX =1234H, Then output AX = 0234H, FX = F002H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 0 1 0 NS NZ NAC PO NC i.e. FL = 02H


Program 4 Memory location Op-code & operand Mnemonics 0000:0100 B8 00 00 MOV AX,0000H 0000:0103 9E SAHF 0000:0104 CD A5 INT A5 Result after execution of above program AX = 0000H, FX = F002H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 0 1 0 NS NZ NAC PO NC 

i.e. FL = 02H

Example 2 : If inputs are AX = FFFFH, Then output AX = FFFFH, FX = F0D7H, D7 D6 D5 D4 D3 D2 D1 D0 1 1 0 1 0 1 1 1 S Z AC PE C i.e. FL = D7H


Example 3 : If inputs are AX =1234H, Then output AX = 1234H, FX = F012H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 1 0 0 1 0 NS NZ AC PO NC i.e. FL = 12H


Program 5 Memory location Op-code & operand Mnemonics 0000:0100 BB 00 00 MOV BX,0000H 0000:0103 B9 00 20 MOV CX,2000H 0000:0106 8E D3 MOV SS,BX 0000:0108 89 CC MOV SP,CX 0000:010A B8 FF FF MOV AX,FFFFH 0000:010D 9C PUSHF 0000:010E 58 POP AX 0000:010F CD A5 INT A5 Result after execution of above program AX = F002H, BX = 0000H, CX = 2000H, SP = 2000H, FX = F002H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 0 1 0 NS NZ NAC PO NC 

i.e. FL = 02H

Example 2 : If 5B POP BX is used Then output BX = F002H, FX = F002H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 0 1 0 NS NZ NAC PO NC 

i.e. FL = 02H

Example 3 : If 59 POP CX is used Then output CX = F002H, FX = F002H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 0 1 0 NS NZ NAC PO NC

i.e. FL = 02H


Program 6 Memory location Op-code & operand Mnemonics 0000:0100 BB 00 00 MOV BX,0000H 0000:0103 B9 00 20 MOV CX,2000H 0000:0106 8E D3 MOV SS,BX 0000:0108 89 CC MOV SP,CX 0000:010A B8 FF FF MOV AX,FFFFH 0000:010D 50 PUSH AX 0000:010E 9D POPF 0000:010F CD A5 INT A5 Result after execution of above program AX = FFFFH, BX = 0000H, CX = 2000H, SP = 2000H, FX = FED7H, D7 D6 D5 D4 D3 D2 D1 D0 1 1 0 1 0 1 1 1 S Z AC PE C 

i.e. FL = D7H

Example 2 : If inputs are AX = 0000H, Then output AX = 0000H, FX = F002H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 0 1 0 NS NZ NAC PO NC 

i.e. FL = 02H

Example 3 : If inputs are AX = 1234H, Then output AX = 1234H, FX = F216H, D7 D6 D5 D4 D3 D2 D1 D0 0 0 0 0 0 1 1 0 NS NZ NAC PE NC

i.e. FL = 16H

Conclusion : The above program and other number of program execution results explain that, after the execution of arithmetic, logical instructions on 8086 microprocessor kit , the flag bits D0,D2,D4,D6 & D7 are changes according to result. But the comparison between LAHF and SAHF instruction, in SAHF execution AH not observed at FL, similarly the comparison between POPF and another POP instruction like POP AX, POP BX etc, in program 6, POPF do not changes accordingly.

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