Assembly Division
assembly, division, x86
Solution
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.
Problem
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!