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Departemen Ilmu Komputer FMIPA IPB 2006. Conditional Jump Instruction A conditional jump instruction transfer control to a destination address when a.

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Presentation on theme: "Departemen Ilmu Komputer FMIPA IPB 2006. Conditional Jump Instruction A conditional jump instruction transfer control to a destination address when a."— Presentation transcript:

1 Departemen Ilmu Komputer FMIPA IPB 2006

2 Conditional Jump Instruction A conditional jump instruction transfer control to a destination address when a flag condition is true Jconddestination Destination address must be -128 to +127 bytes from the current location Cond refer to flag condition, identifying the state of one or more flag

3 ConditionMeaning CCarry flag set NCCarry flag not set (clear) ZZero flag set NZZero flag not set (clear) flags are set by arithmetic, comparison, and boolean instruction. Each conditional jump instruction checks one or more flags, returning a result true or false. If the result is TRUE, otherwise, the program does nothing and continue to the next instruction.

4 cmp ax, bx; compare ax, bx je equal not equal :; continue here if ax <> bx. jmp exit equal :; jump here if ax = bx... exit :; always end up here

5 Case 1 : AX = 5 and BX = 5 Case 2 : AX = 5 and BX = 5

6 Table of Conditional Jump Instruction MnemonicDescriptionFlag Condition JZJump if zeroZF = 1 JEJump if equal (if op1 = op2) JNZJump if not zeroZF = 0 JNEJump if not equal (if op1 <> op2) JAJump if above (if op1 > op2)CF = 0 and ZF = 0 JAEJump if above or equal (if op1 >= op2) CF = 0 JBJump if below (if op1 < op2)CF = 1 JCJump if carry JBEJump if below or equal (if op1 <= op2 ) CF = 1 and ZF = 1 JCXZjump if CX = 0CX = 0

7 Example 1 : larger of two number We might want to compare unsigned value in AX and BX and move the larger one to DX

8 -a A1:0100 mov ax,50 13A1:0103 mov dx,ax 13A1:0105 mov bx,60 13A1:0108 cmp ax,bx 13A1:010A jae 10E 13A1:010C mov dx,bx 13A1:010E int 20 13A1:0110

9 -a A1:0100 mov ax,50 13A1:0103 mov dx,ax 13A1:0105 mov bx,10 13A1:0108 cmp ax,bx 13A1:010A jae 10e 13A1:010C mov dx,bx 13A1:010E int 20 13A1:0110

10 Latihan 1 After the following instruction are executed, what will be the values of AL and BL ? -a A1:0100 mov al,6B 13A1:0102 mov bl,3F 13A1:0104 and ax,0FB6 13A1:0107 cmp al,bl 13A1:0109 ja 10F 13A1:010B mov al,bl 13A1:010D jmp A1:010F mov bl,al 13A1:0111 int 20 13A1:0113

11 Latihan 2 After the following instruction are executed, what will be the values of AL and BL ? -a A1:0100 mov al,3F 13A1:0102 mov bl,6B 13A1:0104 or bl,0F 13A1:0107 sub al,bl 13A1:0109 jb 10F 13A1:010B mov al,1 13A1:010D jmp A1:010F mov bl,1 13A1:0111 int 20 13A1:0113

12 Tugas Buat Program yang Membendingkan 3 bilangan di register al, bl, cl mana yang merupakan bilangan terkecil. Hasilnya di simpan dalam alamat 150, 151, 152 secara berurut mulai dari yang terkecil Nilai awal dari al, bl, cl secara berurut adalah 27, 3, 11 hasil di print screen berserta nama, nama kelompok dan nrp pada alamat 160(nama), 170(nama kelompok), 180 (nrp) Hint : Gunakan fungsi cmp dan jump

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