;;; This program reads an integer limit then counts from 1 to the limt. ;;; Uses plain C I/O and must be linked to the C libraries. global main extern printf extern scanf section .text main: ;; Frame pointer push ebp mov ebp,esp ;; Push registers. Use them as local value, and don't have ;; to worry about functions messing with them. push esi push ebx ;; Issue the prompt. push prompt ; Pushes the address value promt onto the stack call printf ; Call printf with argument prompt add esp,4 ; Removes the argument from the stack. ;; Read the value. push cnt ; Pushes the address value cnt onto the stack. push rfmt ; Ditto rfmt call scanf ; Call scanf with those two arguments. add esp,8 ; Remove them from the stack. ;; Set up for the loop. Init limit and count in registers. mov esi, [cnt] ; Move the value at the address cnt into esi mov ebx,1 ; Move 1 into ebx. ;; Loop test. ltop: cmp ebx,esi ; Compare ebx and esi jg finish ; Jump to finish if ebx > esi (ie count > limit) ;; Print count using the printf function. push ebx ; Push count onto stack (contents of ebx) push pfmt call printf add esp,8 ;; Increment the count inc ebx ; Increment the ebx register. ;; Back to the top jmp ltop ;; Wrap up and leave. finish: pop ebx ; Restore these pop esi mov esp,ebp ; Fix up frame pointer. pop ebp ret ; Go back. section .data prompt: db 'Enter count: ',0 rfmt: db '%d',0 pfmt: db ' => %d',10,0 section .bss cnt: resd 1 ; resd = "Reserve doubleword"