FASM = a * 5 – (b + c + d) / 2

format  PE console
entry   start
include 'win32ax.inc'
include 'api\kernel32.inc'


;----------------------------------------------------------------------
section '.idata' import data readable
library kernel32, 'kernel32.dll',\
        msvcrt,     'msvcrt.dll'

import  msvcrt,\
        printf, 'printf',\
        scanf , 'scanf'



;                                              a * 5 – (b + c + d) / 2
;----------------------------------------------------------------------
section '.data' data readable writeable

           a     dd  1
           b     dd  2
           c     dd  3
           d     dd  4
           r_1   dd  ?     ; = a * 5
           r_2   dd  ?     ; = b + c + d
           r_3   dd  ?     ; = r_2 / 2
           r_4   dd  ?     ; = r_1 - r_3     = a * 5 – (b + c + d) / 2

;----------------------------------------------------------------------
section '.code' code readable executable

start:    call Get_var

          call test_var

          mov eax, [a]      ; r_1 = a * 5
          shl eax, 2
          inc eax
          mov [r_1], eax

          mov eax  , [b]     ; r_2 * a + b + c
          add eax  , [c]
          add eax  , [d]
          mov [r_2], eax

          shr eax, 1         ; r_3 = r_2 / 2
          mov [r_3], eax


          mov eax, [r_1]     ; r-1 - r_3
          sub eax, [r_3]
          mov [r_4], eax

          call test_r_x
          cinvoke  printf, <"         ", 10, 0>
          cinvoke  printf, <"Answer:  ", 10, 0>
          cinvoke  printf, <"- - - - - - - - - - - - - - - - -", 10, 0>
          cinvoke  printf, <"a * 5 - (b + c + d) / 2 = %d", 10, 0>, [r_4]
          ;------------------------
          cinvoke  scanf, <"%d", 10, 0>, 76313
          invoke   ExitProcess, 0






; Функция ожидает от пользователя ввод:  a, b, c, d
; Например, так: 1 22 3 7
;/////////////////////////////////////////////////////////////
proc    Get_var                                            ;//

        cinvoke  printf, <"Enter: a, b, c and d:  ", 0>

        cinvoke  scanf, <"%d%d%d%d">, a, b, c, d
        ret
endp





;Процедура, которая показывает содержимое: a, b, c, d
;----------------------------------------------------------------------
test_var:   cinvoke  printf,           <10, 0>
            cinvoke  printf, <"a = %d", 10, 0>, [a]
            cinvoke  printf, <"b = %d", 10, 0>, [b]
            cinvoke  printf, <"c = %d", 10, 0>, [c]
            cinvoke  printf, <"d = %d", 10, 0>, [d]
ret                                                                    ;Возврат из процедуры

;Процедура, которая показывает содержимое: r_1, r_2, r_3, r_4
;----------------------------------------------------------------------
test_r_x:   cinvoke  printf,           <10, 0>
            cinvoke  printf, <"    a * 5 = r_1 = %d", 10, 0>, [r_1]
            cinvoke  printf, <"b + c + d = r_2 = %d", 10, 0>, [r_2]
            cinvoke  printf, <"  r_2 / 2 = r_3 = %d", 10, 0>, [r_3]
            cinvoke  printf, <"r_1 - r_3 = r_4 = %d", 10, 0>, [r_4]
ret