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I'm working on a program which gets a number (for example, 42 ) then prints it to the screen using interrupt 21 with ah 2. My programs manages to split the number, but when I get a number like 60 my programs calls interrupt 0h because I'm dividing by 0

How can I overcome this?

This is the code:

PROC printNumber push bp push dx push bx push ax mov bp, sp mov ax, [bp + number] mov dx, 0 mov bx, 10 splitNumber: cmp ax, 0 jz exit div bx add dx, '0' mov ah, 2 int 21h jmp splitNumber exit: pop ax pop bx pop dx pop bp retn 2 ENDP printNumber

Thanks! :D

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  • Looks like you forgot to zero DX inside the loop, before the 2nd div. Are you sure this works for 42? Wouldn't the 2nd division run with dx:ax='2':4=0x32:04, leading to divide overflow? Or no, int 21h/ah=2 returns the character printed in AL. spike.scu.edu.au/~barry/interrupts.html#ah02 So that's broken as well. Commented Dec 8, 2019 at 14:43
  • Single-step your code in a debugger to see register-values change Commented Dec 8, 2019 at 14:51
  • @PeterCordes I changed ax and dl to al and dl, but when I'm doing xor dl, dl at the first line of the loop the program prints the first number forever :( Commented Dec 8, 2019 at 14:55
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    div bl (8-bit operand size) writes AH and AL instead of DX and AX. Anyway, it prints the same number forever because (like I commented already), int 21h with AL=2 copies DL to AL, overwriting the rest of the number. Use the debugger built-in to emu8086 Commented Dec 8, 2019 at 15:07
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