EMU8086 dividing 32 bit number by a 16 bit number gives unexpected 0 remainder
问题 I was trying to divide (Unsigned) 8A32F4D5 by C9A5 using emu8086 tool. I expected the quotient to be AF73H and the remainder be 94B6H . After writing the following code, I was getting correct quotient but the remainder became 0000h . Am I missing something? .MODEL SMALL .STACK 100H .DATA .CODE MAIN PROC ; initialize DS MOV AX,@DATA MOV DS,AX ; enter your code here MOV DX, 8A32H MOV AX, 0F4D5H MOV BX, 0C9A5H DIV BX ;exit to DOS MOV AX,4C00H INT 21H MAIN ENDP END MAIN The output in EMU8086: 回答1