程序集指向函数的指针

时间:2017-03-15 20:31:53

标签: assembly fasm

我正在尝试将DWORD变量作为指针参数发送到函数

variable1 dd 1
...
push [variable1]  ; push variable adress
call _InitPoiner
...
_InitPoiner:
    ;
    push ebp
    mov ebp, esp
    ;
    lea eax, [ebp+8]       ; load address
    mov dword [eax], 10    ; move value 10 into that address

    pop ebp
    ret
...
push [variable1]
push sdigit ; where sdigit db '%d', 0x0D, 0x0A, 0
call [printf]

但是variable1是1而不是11,为什么?

2 个答案:

答案 0 :(得分:3)

主要问题是当您执行variable1时,您没有将push [variable1]地址推入堆栈。这会推送存储在variable1的32位值,该值恰好是值1.要推送地址,您将使用push variable1而不使用括号。

在呼叫之前将值推入堆栈时,您需要在呼叫后恢复堆栈。

LEA 没有将地址存储在堆栈中,而是获取应该存储地址的堆栈位置的地址。我想你正在寻找类似的东西:

format ELF

extrn printf
public main

section '.text' executable
main:
    push variable1         ; push address of variable
    call _InitPointer
    add esp, 4             ; We pushed 4 bytes before calling function
                           ; restore the stack by adding 4

    push [variable1]       ; Push the value at address variable1
    push sdigit            ; Format specifier for printf
    call printf
    add esp, 8             ; We pushed 8 bytes of data for call
                           ; restore the stack by adding 8
    ret

_InitPointer:
    push ebp
    mov ebp, esp

    mov eax, [ebp+8]       ; There is an address at [ebp+8] that
                           ; was pushed as the 1st parameter. Put the
                           ; address in EAX

    add dword [eax], 10    ; Add the value 10 to the DWORD value that is at
                           ; address in EAX

    pop ebp
    ret

section '.data' writeable
variable1 dd 1
sdigit db '%d', 0x0D, 0x0A, 0

您没有提及您所使用的平台,因为您没有提供最小的完整可验证示例。上面的代码可以在Linux上测试:

fasm test.asm
gcc -o test -m32 test.o

跑步时你应该得到:

  

11

等效的 C 代码如下所示:

#include <stdio.h>

void _InitPointer (int *ptr)
{
    *ptr += 10;
    return;
}

int variable1 = 1;

int main()
{
    _InitPointer (&variable1);
    printf ("%d\n", variable1);
    return 0;
}

注意:如果在现代版本的Linux上,32位程序遵循System V i386 ABI,要求堆栈在函数调用点处对齐16字节(或32字节)。我在上面的代码中没有做到这一点,以简化它,但应该考虑到它。我不确定你是在Linux还是Windows上。你的问题不说。

32位Windows版本

format PE console 4.0
include 'win32a.inc'

main:
    push variable1         ; push address of variable
    call _InitPointer
    add esp, 4             ; We pushed 4 bytes before calling function
                           ; restore the stack by adding 4

    push [variable1]       ; Push the value at address variable1
    push sdigit            ; Format specifier for printf
    call [printf]
    add esp, 8             ; We pushed 8 bytes of data for call
                           ; restore the stack by adding 8
    push 0                 ; Exit with return value 0
    call [ExitProcess]

_InitPointer:
    push ebp
    mov ebp, esp

    mov eax, [ebp+8]       ; There is an address at [ebp+8] that
                           ; was pushed as the 1st parameter. Put the
                           ; address in EAX

    add dword [eax], 10    ; Add the value 10 to the DWORD value that is at
                           ; address in EAX

    pop ebp
    ret

variable1 dd 1
sdigit db '%d', 0x0D, 0x0A, 0

data import
 library kernel32,'KERNEL32.DLL',\
         msvcrt,'MSVCRT.DLL'
 import msvcrt,\
        printf ,'printf'
 import kernel32,\
        ExitProcess,'ExitProcess'
end data

答案 1 :(得分:2)

你确定在完成后弹出你的变种?

看看你的例子,我认为变量不可能是11.它在dd赋值时从1开始,然后如果你在lea中的数学是正确的,那么它将是10。您可以通过调试器执行此操作,您可以检查您的lea / mov组合是否正常工作。无论哪种方式,我都期望1或10,而不是11。

也许您打算add代替mov