;;; 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"