; call2.asm  code loopint.c for nasm 
; /* call2.c a very simple loop that will be coded for nasm */
; #include <stdio.h>
; static void call2(int *A, int start, int end, int value);
; int main()
; {
;   int dd1[100];
;   int i;
;   dd1[0]=5; /* be sure loop stays 1..98 */
;   dd1[99]=9;
;   call2(dd1,1,98,7); /* fill dd1[1] thru dd1[98] with 7 */
;   printf("dd1[0]=%d, dd1[1]=%d, dd1[98]=%d, dd1[99]=%d\n",
;           dd1[0], dd1[1], dd1[98],dd1[99]);
;   return 0;
; }
; static void call2(int *A, int start, int end, int value)
; {
;   int i;
;
;   for(i=start; i<=end; i++) A[i]=value;
; }

; execution output is dd1[0]=5, dd1[1]=7, dd1[98]=7, dd1[99]=9
 
	section	.bss
i:	resd	1		; actually unused, kept in register

	section .text
	global	call2		; linker must know name of subroutine
call2:				; name must appear as a nasm label
        push	ebp		; save ebp
        mov	ebp, esp	; ebp is callers stack
        push	ebx		; save registers
        push	edi

        mov	edi,[ebp+8]	; get address of A into edi
        mov	eax,[ebp+12]	; get value of start
        mov	ebx,[ebp+16]	; get value of end
        mov	edx,[ebp+20]	; get value of value
	
loop1:	mov	[4*eax+edi],edx	; A[i]=value;
	add	eax,1		; i++;
	cmp	eax,ebx		; i<=end
	jle	loop1		; loop until i=9

	pop	edi		; in reverse order
        pop	ebx
        mov	esp,ebp		; restore callers stack frame
        pop	ebp
        ret			; return
;
; Notes about the callers stack, ebp in our code:
; ebp+8  is the last argument passed to us by the caller,
;        this is our first argument, an address.
; ebp+12 would is our second argument, start a value.