如何在Assembly Works中调用带有大量args的函数

Question

如何在Assembly Works中调用带有大量args的函数

Lan*_*ard 1 c linux assembly x86-64 function

这是一组示例函数,第一个是20个args,第二个是2:

int a(int n1, int n2, int n3, int n4, int n5, int n6, int n7, int n8, int n9, int n10, int n11, int n12, int n13, int n14, int n15, int n16, int n17, int n18, int n19, int n20) {
    return n1 * n2 * n3 * n4 * n5 * n6 * n7 * n8 * n9 * n10 * n11 * n12 * n13 * n14 * n15 * n16 * n17 * n18 * n19 * n20;
}

int b(int n1, int n2) {
    return a(n1, n2, n1, n2, n1, n1, n2, n1, n2, n1, n1, n2, n1, n2, n1, n1, n2, n1, n2, n1)
      + a(n1, n2, n1, n2, n1, n1, n2, n1, n2, n1, n1, n2, n1, n2, n1, n1, n2, n1, n2, n1)
      + a(n1, n2, n1, n2, n1, n1, n2, n1, n2, n1, n1, n2, n1, n2, n1, n1, n2, n1, n2, n1);
}

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它被编译到这个程序集:

a(int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int):
  push    rbp
  mov     rbp, rsp
  mov     DWORD PTR [rbp-4], edi
  mov     DWORD PTR [rbp-8], esi
  mov     DWORD PTR [rbp-12], edx
  mov     DWORD PTR [rbp-16], ecx
  mov     DWORD PTR [rbp-20], r8d
  mov     DWORD PTR [rbp-24], r9d
  mov     eax, DWORD PTR [rbp-4]
  imul    eax, DWORD PTR [rbp-8]
  imul    eax, DWORD PTR [rbp-12]
  imul    eax, DWORD PTR [rbp-16]
  imul    eax, DWORD PTR [rbp-20]
  imul    eax, DWORD PTR [rbp-24]
  imul    eax, DWORD PTR [rbp+16]
  imul    eax, DWORD PTR [rbp+24]
  imul    eax, DWORD PTR [rbp+32]
  imul    eax, DWORD PTR [rbp+40]
  imul    eax, DWORD PTR [rbp+48]
  imul    eax, DWORD PTR [rbp+56]
  imul    eax, DWORD PTR [rbp+64]
  imul    eax, DWORD PTR [rbp+72]
  imul    eax, DWORD PTR [rbp+80]
  imul    eax, DWORD PTR [rbp+88]
  imul    eax, DWORD PTR [rbp+96]
  imul    eax, DWORD PTR [rbp+104]
  imul    eax, DWORD PTR [rbp+112]
  imul    eax, DWORD PTR [rbp+120]
  pop     rbp
  ret
b(int, int):
  push    rbp
  mov     rbp, rsp
  push    rbx
  sub     rsp, 8
  mov     DWORD PTR [rbp-12], edi
  mov     DWORD PTR [rbp-16], esi
  mov     r9d, DWORD PTR [rbp-12]
  mov     r8d, DWORD PTR [rbp-12]
  mov     ecx, DWORD PTR [rbp-16]
  mov     edx, DWORD PTR [rbp-12]
  mov     esi, DWORD PTR [rbp-16]
  mov     eax, DWORD PTR [rbp-12]
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, eax
  call    a(int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int)
  add     rsp, 112
  mov     ebx, eax
  mov     r9d, DWORD PTR [rbp-12]
  mov     r8d, DWORD PTR [rbp-12]
  mov     ecx, DWORD PTR [rbp-16]
  mov     edx, DWORD PTR [rbp-12]
  mov     esi, DWORD PTR [rbp-16]
  mov     eax, DWORD PTR [rbp-12]
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, eax
  call    a(int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int)
  add     rsp, 112
  add     ebx, eax
  mov     r9d, DWORD PTR [rbp-12]
  mov     r8d, DWORD PTR [rbp-12]
  mov     ecx, DWORD PTR [rbp-16]
  mov     edx, DWORD PTR [rbp-12]
  mov     esi, DWORD PTR [rbp-16]
  mov     eax, DWORD PTR [rbp-12]
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, DWORD PTR [rbp-12]
  push    rdi
  mov     edi, DWORD PTR [rbp-16]
  push    rdi
  mov     edi, eax
  call    a(int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int, int)
  add     rsp, 112
  add     eax, ebx
  mov     rbx, QWORD PTR [rbp-8]
  leave
  ret

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我有几个问题.首先,我注意到它似乎随着数量的增加而切换它如何处理args:

push    rbp
mov     rbp, rsp
mov     DWORD PTR [rbp-4], edi
...
imul    eax, DWORD PTR [rbp-8]
...

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想知道那里发生了什么,为什么会这样做.它似乎对待第一个arg push,然后接下来的8个左右mov,然后剩下的imul仅与eax寄存器有关.想知道你可以拥有多少个args是否有限制.

我想知道的第二件事是以下内容.比如说a(),该函数b()调用了一些"系统函数"或其他一些外部库调用.想知道它是如何解开args的.嗯,我想,没关系,我假设C编译器将编译成汇编/机器代码所有链接的外部库.所以,我猜是的.

最后一件事是,如果系统调用都有一定数量的参数,类似于x86中最大操作数为3的方式.或者系统调用可能有任意数量的参数.看来他们想限制它的性能,所以他们只使用那些早期指示像push和mov的,而不是imul.

感谢您的帮助,只需查看在汇编时调用函数时如何处理参数的说明,尤其是在存在大量参数时.

Answer 1

小智 5

您需要了解有关堆栈帧和应用程序二进制接口(ABI或调用约定)的更多信息.ABI定义调用者将参数传递给被调用者的方式,寄存器是易失性的,以及如何清理堆栈.

存在许多ABI,因为只要呼叫者和被呼叫者同意,任何人都可以设计他们自己的ABI.但是,只有少数ABI被广泛使用.在Windows上,大多数32位程序使用stdcall,cdecl,Microsoft的fastcall或Borland的fastcall,而64位程序大多使用Microsoft x64调用约定.在Unix 64位程序上总是使用System V AMD64 ABI,它也是编译器使用的.

让我们看一下您的代码,并附上评论:

push    rbp                     ; save the old stack frame
mov     rbp, rsp                ; establish new stack frame
mov     DWORD PTR [rbp-4], edi  ; save the first six arguments
mov     DWORD PTR [rbp-8], esi
mov     DWORD PTR [rbp-12], edx
mov     DWORD PTR [rbp-16], ecx
mov     DWORD PTR [rbp-20], r8d
mov     DWORD PTR [rbp-24], r9d
mov     eax, DWORD PTR [rbp-4]  ; load n1
imul    eax, DWORD PTR [rbp-8]  ; eax = eax * n2
imul    eax, DWORD PTR [rbp-12] ; eax = eax * n3
imul    eax, DWORD PTR [rbp-16]
imul    eax, DWORD PTR [rbp-20]
imul    eax, DWORD PTR [rbp-24]
imul    eax, DWORD PTR [rbp+16] ; eax = eax * n7
imul    eax, DWORD PTR [rbp+24] ; eax = eax * n8
imul    eax, DWORD PTR [rbp+32]
imul    eax, DWORD PTR [rbp+40]
imul    eax, DWORD PTR [rbp+48]
imul    eax, DWORD PTR [rbp+56]
imul    eax, DWORD PTR [rbp+64]
imul    eax, DWORD PTR [rbp+72]
imul    eax, DWORD PTR [rbp+80]
imul    eax, DWORD PTR [rbp+88]
imul    eax, DWORD PTR [rbp+96]
imul    eax, DWORD PTR [rbp+104]
imul    eax, DWORD PTR [rbp+112]
imul    eax, DWORD PTR [rbp+120]
pop     rbp                      ; restore old stack frame
ret                              ; exit

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注意:前两行与参数无关; 它们是创建一个堆栈框架,以便您可以轻松访问局部变量和参数.如果没有堆栈帧你仍然可以用[RSP +*]访问它们,但偏移根据需要任意调整PUSH和POP您使用.

接下来是将参数存储到局部变量的指令.寄存器经常被更改,并且需要存储寄存器中传递的参数,以防您以后需要使用它们.但是,在这种情况下,没有必要.所以优化的代码可以

push    rbp                     ; save the old stack frame
mov     rbp, rsp                ; establish new stack frame
mov     eax, edi                ; eax = n1
imul    eax, esi                ; eax = eax * n2
imul    eax, edx                ; eax = eax * n3
imul    eax, ecx                ; eax = eax * n4
imul    eax, e8d                ; eax = eax * n5
imul    eax, e9d                ; eax = eax * n6
imul    eax, DWORD PTR [rbp+16] ; eax = eax * n7
imul    eax, DWORD PTR [rbp+24] ; eax = eax * n8
imul    eax, DWORD PTR [rbp+32] ; eax = eax * n9
imul    eax, DWORD PTR [rbp+40]
imul    eax, DWORD PTR [rbp+48]
imul    eax, DWORD PTR [rbp+56]
imul    eax, DWORD PTR [rbp+64]
imul    eax, DWORD PTR [rbp+72]
imul    eax, DWORD PTR [rbp+80]
imul    eax, DWORD PTR [rbp+88]
imul    eax, DWORD PTR [rbp+96]
imul    eax, DWORD PTR [rbp+104]
imul    eax, DWORD PTR [rbp+112]
imul    eax, DWORD PTR [rbp+120]
pop     rbp                      ; restore old stack frame
ret                              ; exit

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从上面的例子,你应该能够猜出的第一个参数是在传递edi(或rdi,di,dil取决于大小),第二个是在esi,然后edx,ecx,r8d和r9d(整数只,浮标在矢量寄存器中传递) .当你有超过6个参数时,另一个被推到堆栈并且可以使用[rbp + 16],[rbp + 24],...来访问([rbp + 8]是旧的rbp; [rbp]正在返回地址).

对于来电者

mov     r9d, DWORD PTR [rbp-12]  ; r9d = n6
mov     r8d, DWORD PTR [rbp-12]  ; r8d = n5
mov     ecx, DWORD PTR [rbp-16]  ; ecx = n4
mov     edx, DWORD PTR [rbp-12]  ; edx = n3
mov     esi, DWORD PTR [rbp-16]  ; esi = n2
mov     eax, DWORD PTR [rbp-12]  ; eax = n1  ; will assign to edi
mov     edi, DWORD PTR [rbp-12]  ; push n20
push    rdi
mov     edi, DWORD PTR [rbp-16]  ; push n19
push    rdi
mov     edi, DWORD PTR [rbp-12]
push    rdi
mov     edi, DWORD PTR [rbp-16]
push    rdi
mov     edi, DWORD PTR [rbp-12]
push    rdi
mov     edi, DWORD PTR [rbp-12]
push    rdi
mov     edi, DWORD PTR [rbp-16]
push    rdi
mov     edi, DWORD PTR [rbp-12]
push    rdi
mov     edi, DWORD PTR [rbp-16]
push    rdi
mov     edi, DWORD PTR [rbp-12]
push    rdi
mov     edi, DWORD PTR [rbp-12]
push    rdi
mov     edi, DWORD PTR [rbp-16]
push    rdi
mov     edi, DWORD PTR [rbp-12]
push    rdi
mov     edi, DWORD PTR [rbp-16]  ; push n7
push    rdi
mov     edi, eax                 ; edi = n1
call    a()                      ; call the function
add     rsp, 112                 ; clean up the stack, 14 * 8 = 112 bytes
mov     ebx, eax                 ; result is in eax

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一个更直接的版本是

mov     r9d, DWORD PTR [rbp-12]  ; r9d = n6
mov     r8d, DWORD PTR [rbp-12]  ; r8d = n5
mov     ecx, DWORD PTR [rbp-16]  ; ecx = n4
mov     edx, DWORD PTR [rbp-12]  ; edx = n3
mov     esi, DWORD PTR [rbp-16]  ; esi = n2
mov     edi, DWORD PTR [rbp-12]  ; edi = n1
push    [rbp-12]                 ; push n20
push    [rbp-16]                 ; push n19
push    [rbp-12]
push    [rbp-16]
push    [rbp-12]
push    [rbp-16]
push    [rbp-12]
push    [rbp-16]
push    [rbp-12]
push    [rbp-16]
push    [rbp-12]
push    [rbp-16]
push    [rbp-12]                 ; push n8
push    [rbp-16]                 ; push n7
call    a()                      ; call the function
add     rsp, 112                 ; clean up the stack, 14 * 8 = 112 bytes
mov     ebx, eax                 ; result is in eax

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请注意,参数按相反顺序推送.

由于您可以在调用函数之前将任何数字推送到堆栈(在它溢出之前),因此对参数的数量没有限制.

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