Question

please provide screenshots using dos debug.

4. Assemble the following instruction sequence into the memory starting at address CS: 100 and then verify their machine code in the memory. a. ADD AX, 00FFH b. ADC SI, AX c. INC BYTE PTR [0100H] d. SUB DL, BL e. SBB DL, [0200H] f. DEC BYTE PTR [DI+BX] g. NEG BYTE PTR [DIJ+0010H h. MUL DX i. IMUL WORD PTR [BX+SI] j. DIV WORD PTR [SI]+0030H k. IDIV WORD PTR [BX]ISI]+0030HS. How many bytes of memory do the instructions take up? ay encounter an interrupt/exception. Try to find the reason behind Note: For instructions j and k, you it, and modify the contents of registers with new values to avoid the divide overflow.

Answer 1

Here is my screenshot of DOS Debug window:

a 100 072A:0100 add ax,0off 072A:0103 adc si,ax 072A:0105 inc byte ptrI01001 072A:0109 sub dl,bl 072A:01OB sbb d1, 02001 072A:010F dec byte ptrldi+bxl 072A:0111 neg byte ptrIdil+0010 Error 072A:0111 mul dx 072A:0113 imul word ptrbx+sil 072A:0115 div word ptr[sil+0030 Error OP2A:0115 idiv word ptrlbx]Isi l+0030h5 Error 072A:0115

here 3 instruction are giving error they are,

neg byte ptr[di]+0010

div word ptr[si]+0030

idiv word ptr[bx][si]+0030h5

above this instructions are giving error because in those instruction addressing mode is wrong.

We can not use addressing mode like [di]+0010.So due that addressing mode these three instructions give error.

Now we calculate byte of memory each instruction take up,

1) 0100 - 0103 = 3 in HEX = 3 in decimal = 3 bytes

2) 0105 - 0103 = 2 bytes

3) 0109 - 0105 = 4 bytes

4) 010B - 0109 = 2 bytes

5) 010F - 010B = 4 bytes

6) 0111 - 010F = 2 bytes

7) 0113 - 0111 = 2 bytes      //for mul dx

8) 0115 - 0113 = 2 bytes      //for imul word ptr[bx+si]

Now We have discussed that three instructions give error so that we can not calculate how memory bytes it take up,but i directly say that how many bytes they take up if they are modified like,

For neg byte ptr[di]+0010 if we change it to neg byte ptr[di+0100] then this will word and it takes 3 bytes.

For div word ptr[si]+0030 change it to div word ptr[si+0030] then it will take 3 bytes.

For idiv word ptr[bx][si]+0030h5 change it to idiv word ptr[bx][si+0030] then it will also take 3 bytes.

if you have any query then feel free to ask in comment.