获取过程参数从堆栈 [英] Getting Procedure Arguments from Stack

问题描述

我想学习如何调用汇编语言程序。下面是一个简单的例子，显示我的问题。我推 7 对栈，调用程序;当procudeure从堆栈中弹出，这个值是不是我推的人。是否有人可以帮助我明白发生了什么，什么我可以做，使这项工作？

  PUSH 7
CALL FOOBAR
MOV AH，4CH
INT 21HFOOBAR PROC
    POP AX;没有7
    RET
FOOBAR ENDP
 

解决方案

的呼叫指令将返回地址在堆栈上，所以当你流行斧在你的程序，它不会让你7推，但返回地址。在 RET 将无法正常工作，无论是（它希望找到返回地址有！）试着这么做...

  FOOBAR PROC
推基点;保存调用者的章
MOV BP，SP
MOV AX，[BP + 4]
;用它做什么
;离开 - 相当于：
MOV SP，BP
流行基点RET
 

有一个可能的疑难杂症在这里。 A远的过程既​​有段（CS）和堆栈上的偏移，所以4轮空，返回地址，并为推BP 两个字节将第一个参数在 [BP + 6] 。我猜只是 PROC 默认为附近 PROC - 你可能想要说的是，仅仅是为了清楚。如果你需要一个 PROC远它可能是时候升级到32位code（或64位）。 16位code是这样的皮塔饼 - 我们真的很高兴算了吧！ ：）

I am trying to learn how to call procedures in assembly language. Below is a simple example that shows my problem. I push 7 unto the stack, call the procedure; when the procudeure pops from the stack, the value is not the one I pushed. Can someone please help me understand what is happening and what I can do to make this work?

PUSH 7
CALL FOOBAR
MOV AH, 4CH
INT 21H

FOOBAR PROC
    POP AX ; Not 7
    RET
FOOBAR ENDP

解决方案

The call instruction puts the return address on the stack, so when you pop ax in your procedure, it doesn't get the 7 you pushed, but the return address. The ret won't work, either (it expects to find the return address there!) try something like...

FOOBAR proc
push bp ; save caller's reg
mov bp, sp
mov ax, [bp + 4]
; do something with it
; leave - equivalent to:
mov sp, bp
pop bp

ret

There's a possible "gotcha" here. A "far" procedure has both the segment (cs) and offset on the stack, so 4 byes for the return address, and two bytes for the push bp puts the first parameter at [bp + 6]. I guess just proc defaults to proc near - you might want to say that, just for clarity. If you need a proc far it's probably time to graduate to 32-bit code (or 64-bit). 16-bit code is such a PITA - we're really glad to forget it! :)

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