EBP，ESP和在NASM汇编栈帧 [英] Ebp, esp and stack frame in assembly with nasm

问题描述

我对EBP的一些问题，尤其在以下code堆栈帧。

1. 为什么我们从。减去28 ESP？我们有两个主要的局部变量x和y。那么为什么我们没有8。减去




2. 和我们不把值从右堆到剩下什么？那么，为什么我们增加1到[EAX + 8]而不是[EAX + 4]？




3. 我感到有点混淆这个结构。你能帮我吗？ THX。

     FUNC（INT A，INT B，INT C）
   {
     返回+ B + C;
   }
   主要（）
   {
    INT X，Y = 3;
    X = FUNC（Y，2,1）;
   }
    

解决方案

1. 该堆栈指针28，因为你需要8个字节为你的两个局部变量和参数，以FUNC 12字节中减去。额外的8个字节可能是由于你的编译器的企图对准主要的筹码为16字节边界（还有已经在栈上的4个字节为主要的返回地址和另外4个字节时EBP被推建立在主的第一个指令堆栈帧）。见<一href=\"http://gcc.gnu.org/onlinedocs/gcc-4.2.0/gcc/i386-and-x86_002d64-Options.html\">-m$p$pferred-stack-boundary如果你使用GCC。




2. 正在传递的参数从右到左。由于堆栈空间已经分配给三个参数，当它被从堆栈指针减去1被移动到相对于当前堆栈指针（+8）的最高的位置，2被移动到中间（+4）和y中的值被移动到堆栈指针本身。这是相同的堆栈上推1，在堆栈上推2，然后在栈上推年。由最后推指令，图1是从ESP +8，2是从ESP + 4，y是从ESP + 0。注意FUNC的里面，它有增加8到这些偏移，因为返回地址被压从调用指令堆栈和FUNC EBP推动建立一个堆栈帧。




3. 混淆有关的结构？

I have a few questions about ebp, esp and stack frame in following code.

1. Why did we substract 28 from esp ? We have two local variables x and y in main. So why didn't we substract 8?

2. And don't we put values to stack from right to left? So why did we add 1 to [eax+8] instead of [eax+4] ?

3. I am a little bit confuse about this structure. Can you help me out? Thx.

   func(int a, int b, int c)
   {
     return a+b+c;
   }
   main()
   {
    int x, y=3;
    x=func(y,2,1);
   }
   

解决方案

1. The stack pointer is subtracted by 28 because you need 8 bytes for your two local variables and 12 bytes for the parameters to func. The extra 8 bytes are likely due to your compiler's attempt to align main's stack to a 16-byte boundary (there's already 4 bytes on the stack for main's return address and another 4 bytes when EBP was pushed to establish the stack frame in main's first instruction). See -mpreferred-stack-boundary if you're using GCC.

2. The parameters are being passed right-to-left. Since the stack space was already allocated for the three parameters when it was subtracted from the stack pointer, 1 is moved into the "highest" position relative to the current stack pointer (+8), 2 is moved into the middle (+4), and the value in y is moved into the stack pointer itself. This is the same as pushing 1 on the stack, pushing 2 on the stack, and then pushing y on the stack. By the last push instruction, 1 is +8 from ESP, 2 is +4 from ESP, and y is +0 from ESP. Note that inside of func, it has to add 8 to these offsets because the return address is pushed on the stack from the call instruction and func pushes EBP to establish a stack frame.

3. Confused about which structure?

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