调用glibc的时候x86-64的ELF初始堆栈布局 [英] x86-64 ELF initial stack layout when calling glibc

问题描述

基本上，我通过 http://www.nasm.us/links/unix64abi，并在第29页，它显示了一个C程序的初始进程堆栈。

我的问题是：我试图从X86-64 NASM用glibc接口，并根据上面显示，ARGC应该在RSP什么。所以下面code应打印ARGC：

  [段.data]
PrintStr：DB你刚刚踏入％d个参数。，10，0[.bss段][.text段]
EXTERN的printf
全球主要主要：
     MOV RAX，0;需要采取的变量没有的功能。 args来
     MOV RDI，PrintStr
     RSI MOV [RSP]
     调用printf
     RET
 

但事实并非如此。有人可以告诉我，如果我在code犯任何错误或告诉我实际的堆栈结构是什么？

谢谢！

更新：我只是随意尝试了一些偏移和改变MOV RSI，[RSP]到MOV RSI，[RSP + 28]的伎俩

但是，这意味着，所示的堆结构体是错误的。有谁知道初始堆栈布局是一个小精灵x86-64的？ http://asm.sourceforge.net/articles/startup.html 的等效会是非常好的。

更新2：
我离开了我如何建立这个code。我通过做：

  NASM -f ELF64 -g＆LT;＆名GT;
GCC＆LT;＆名GT;的.o -o＆LT;＆OUTPUTFILE GT;
 

解决方案

初始堆栈布局包含 ARGC 堆栈指针，随后的阵列 的char * argv的[] ，而不是一个指针，它像主接收。因此，打电话给主，你需要做的是这样的：

 弹出％RDI
MOV％RSP，RSI％
调用主
 

在现实中通常调用主，而不是启动code直接这样做。

的包装函数

如果你想简单地打印的argv [0] ，你可以这样做：

 弹出％RDI
流行％RDI
电话看跌期权
XOR％EDI，EDI％
JMP出口
 

Basically, I read through parts of http://www.nasm.us/links/unix64abi and at page 29, it shows the initial process stack of a C program.

My question is: I'm trying to interface with glibc from x86-64 nasm and based on what the above shows, argc should be at rsp. So the following code should print argc:

[SECTION .data]
PrintStr: db "You just entered %d arguments.", 10, 0

[SECTION .bss]

[SECTION .text]
extern printf
global main

main:
     mov rax, 0        ; Required for functions taking in variable no. of args
     mov rdi, PrintStr
     mov rsi, [rsp]
     call printf
     ret

But it doesn't. Can someone enlighten me if I have made any mistakes in my code or tell me what the actual stack structure is?

Thanks!

UPDATE: I just randomly tried some offsets and changing the "mov rsi, [rsp]" to "mov rsi, [rsp+28]" did the trick.

But this means that the stack structure shown is wrong. Does anyone know what the initial stack layout is for an x86-64 elf? An equivalent of http://asm.sourceforge.net/articles/startup.html would be really nice.

UPDATE 2: I left out how I build this code. I do it by:

nasm -f elf64 -g <filename>
gcc <filename>.o -o <outputfile>

解决方案

The initial stack layout contains argc at the stack pointer, followed by the array char *argv[], not a pointer to it like main receives. Therefore, to call main, you need to do something like:

pop %rdi
mov %rsp,%rsi
call main

In reality there is usually a wrapper function that calls main, rather than the startup code doing it directly.

If you want to simply print argv[0], you could do something like:

pop %rdi
pop %rdi
call puts
xor %edi,%edi
jmp exit

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