sub3.asm

关于linux环境下的nasm代码的生成和使用

;
; file: sub3.asm
; Subprogram example program
;
; To create executable:
; Using djgpp:
; nasm -f coff sub3.asm
; gcc -o sub1 sub3.o driver.c asm_io.o
;
; Using Borland C/C++
; nasm -f obj sub3.asm
; bcc32 sub3.obj driver.c asm_io.obj

%include "asm_io.inc"

segment .data
sum     dd   0

segment .bss
input   resd 1

 

;
; psuedo-code algorithm
; i = 1;
; sum = 0;
; while( get_int(i, &input), input != 0 ) {
;   sum += input;
;   i++;
; }
; print_sum(num);

segment .text
        global  asm_main
asm_main:
        enter   0,0               ; setup routine
        pusha

        mov     edx, 1            ; edx is 'i' in pseudo-code
while_loop:
        push    edx               ; save i on stack
        push    dword input       ; push address on input on stack
        call    get_int
        add     esp, 8            ; remove i and &input from stack

        mov     eax, [input]
        cmp     eax, 0
        je      end_while

        add     [sum], eax        ; sum += input

        inc     edx
        jmp     short while_loop

end_while:
        push    dword [sum]       ; push value of sum onto stack
        call    print_sum
        pop     ecx               ; remove [sum] from stack

        popa
        leave                     
        ret

;
; subprogram get_int
; Parameters (in order pushed on stack)
;   number of input (at [ebp + 12])
;   address of word to store input into (at [ebp + 8])
; Notes:
;   values of eax and ebx are destroyed
segment .data
prompt  db      ") Enter an integer number (0 to quit): ", 0

segment .text
get_int:
        push    ebp
        mov     ebp, esp

        mov     eax, [ebp + 12]
        call    print_int

        mov     eax, prompt
        call    print_string
        
        call    read_int
        mov     ebx, [ebp + 8]
        mov     [ebx], eax         ; store input into memory

        pop     ebp
        ret                        ; jump back to caller

; subprogram print_sum
; prints out the sum
; Parameter:
;   sum to print out (at [ebp+8])
; Note: destroys value of eax
;
segment .data
result  db      "The sum is ", 0

segment .text
print_sum:
        push    ebp
        mov     ebp, esp

        mov     eax, result
        call    print_string

        mov     eax, [ebp+8]
        call    print_int
        call    print_nl

        pop     ebp
        ret