org $100 ;program starting location 100h

;mov esp,$FABC

push esi edx ecx ebx eax ; put registers on stack
mov si,axeq ; display variable offset
call bwritestring ; put variable at offset si on screen
mov ebx,16 ; hexadecimal display of value
pop eax ; load eax from stack
call putlongint ; display value in eax
call bendline   ; end the line

mov si,bxeq
call bwritestring
pop eax
call putlongint
call bendline

mov si,cxeq
call bwritestring
pop eax
call putlongint
call bendline

mov si,dxeq
call bwritestring
pop eax
call putlongint
call bendline

mov si,cseq
call bwritestring
mov ax,cs
call putlongint
call bendline

mov si,dseq
call bwritestring
mov ax,ds
call putlongint
call bendline

mov si,eseq
call bwritestring
mov ax,es
call putlongint
call bendline

mov si,fseq
call bwritestring
mov ax,fs
call putlongint
call bendline

mov si,gseq
call bwritestring
mov ax,gs
call putlongint
call bendline

mov si,sieq
call bwritestring
pop eax
call putlongint
call bendline

mov si,dieq
call bwritestring
mov eax,edi
call putlongint
call bendline

mov si,sseq
call bwritestring
mov ax,ss
call putlongint
call bendline

push dword $12345678
;pop eax

mov si,speq
call bwritestring
mov bx,16
mov eax,esp
call putlongint
call bendline

mov si,spdeq
call bwritestring
mov edi,esp
;mov eax,[edi-4] ;- esp-2 cimen sp értéke van
mov eax,[edi]
call putlongint
call bendline

pop eax

mov si,bpeq
call bwritestring
mov eax,ebp
call putlongint
call bendline


cli             ; clear interrupts while we setup a stack
mov ax,0x9000   ; this seems to be the typical place for a stack
mov ss,ax
mov sp,0xffff   ; let's use the whole segment.  Why not?  We can 
sti             ; put our interrupts back on


int 20h ; exit function there are alternatives

axeq db 'EAX=$',0 ; variable declarations.- messages to display
bxeq db 'EBX=$',0
cxeq db 'ECX=$',0
dxeq db 'EDX=$',0
cseq db 'CS=$',0
dseq db 'DS=$',0
eseq db 'ES=$',0
fseq db 'FS=$',0
gseq db 'GS=$',0
sieq db 'ESI=$',0
dieq db 'EDI=$',0
sseq db 'SS=$',0
speq db 'ESP=$',0
spdeq db '[ESP]=$',0
bpeq db 'EBP=$',0

bendline: ; ends the line ( adds y loc = 1 x loc = 0 )
push ax bx
mov bx,7 ; returns:  AX = 0xE0A ; BX = 7
mov ax,$E0D
int 10h
mov al,$A
int 10h
pop bx ax
ret

bwritestring: ; Writes a 0 terminated string to screen ( string is at ds:si )
push ax bx si
mov bx,0007h ; use video page 0, normal white
localloop:
lodsb
or al,al
jnz next_char
pop si bx ax
ret
next_char:
mov ah,0eh
int 10h
jmp localloop

putint: ; AX = number, BX = base
  push ax bx cx dx
  cmp bx,2 ; base can't be less than 2 
  jge .start
  mov bx,10     ; using bx = 10 instead
  .start:
  xor cx,cx     ; cx = 0
  .new:
  xor dx,dx     ; dx = 0
  div bx        ; number / base
  push dx       ; push the remainder
  inc cx        ; increase the "digit-count"
  or ax,ax      ; if the quotient still is not 0, do it once more
  jnz .new
  .loop:
  pop ax        ; pop the remainder
cmp al,10
sbb al,69h
das
push bx
mov bx,15 ;mov bl,15 ;xor bh,bh
mov ah,$e
int 10h
pop bx
  loop .loop
  pop dx cx bx ax
ret

putlongint: ; EAX = number, EBX = base
  push eax ebx ecx edx
  cmp ebx,2 ; base can't be less than 2 
  jge .start
  mov ebx,10     ; using bx = 10 instead
  .start:
  xor ecx,ecx     ; cx = 0
  .new:
  xor edx,edx     ; dx = 0
  div ebx        ; number / base
  push dx       ; push the remainder
  inc ecx        ; increase the "digit-count"
  or eax,eax      ; if the quotient still is not 0, do it once more
  jnz .new
  .loop:
  pop ax        ; pop the remainder
cmp al,10
sbb al,69h
das
push bx
mov bx,15 ;mov bl,15 ;xor bh,bh
mov ah,$e
int 10h
pop bx
  loop .loop
  pop edx ecx ebx eax
ret
    

For FASM, because a TASM tutorial does no good with FASM

org 100h ; start

; code here

ret ; int 20h ;... exit

; variables
    

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