Assembly Division [duplicate]

Question

This question already has an answer here:

• Why should EDX be 0 before using the DIV instruction? 2 answers

In my program, a hex number is divided by ten and the remainder is checked.

First division is performed well; however, after the second division, the program goes wrong. I am new to assembly, and I couldn't find where the problem is...

Here is the code segment:

ORG 1000

    MOV AX, 0x04B4 (1204 decimal value )
    MOV BX, 0x000A ( 10 decimal value )
    MOV CX, 0x0000

    DIV BX ( After this part, AX is 120 decimal and DX 4 decimal )

    CMP DX, 0x0000
    JE eq1

    ADD CX, 0x0002
    JMP con1

    eq1:    ADD CX, 0x0001  

    con1:

    DIV BX ( But, after this division AX becomes 6677 ( 26231 decimal and DX remains 4 decimal )
    CMP DX, 0x0000

Thanks for help!

Answer 1

The DIV BX instruction divides the 32-bit value in DX:AX by BX. Since you're not initializing DX, the upper word of the dividend is whatever garbage was left in the DX register from the previous computation, so you're really dividing 0x00040078=262314 by 10. The result is correct: a quotient of 26231 with a remainder of 4.

In the first division is must have been pure luck that DX happened to be 0 initially.

Answer 2

Intel instruction DIV divides the register pair DX:AX with the argument.

In the first case DX happens to be zero.
The second time DX must have been 4.