原文

http://kakaroto.homelinux.net/2017/11/introduction-to-reverse-engineering-and-assembly/

翻译部分段落

1. Stack

Stack

栈内存储数据

上图展示了栈的LIFO（后进先出）性质，它在RAM中是向下增长（和图中向上增加是相反的）。栈用来存储局部变量和函数的返回地址（来自上一层函数调用该函数后的指令）。

一个栈里有多个栈帧，可以看到当前的栈帧，包括了它所有的变量，和调用它的函数的返回地址，在它之上，是上一个函数的帧，也是包含了它自己的变量和调用它的函数的返回地址，依此类推，main函数在栈的顶端。

2. Registers

寄存器介绍

可以把一个寄存器当作一个变量，x86处理器上一共只有9个寄存器，其中只有7个是可用的。
x86上的变量寄存器是这些：EAX, EBX, ECX, EDX, EDI, ESI, EBP, ESP, EIP.

其中两个比较特殊：
EIP：指令指针，存储当前正在执行的指令的地址；
ESP：栈指针，存储当前栈的地址。

3. Instructions

常用指令

Call指令

调用一个函数。
相当于两步：

PUSH %EIP+4
JMP

Sample

int main() {
return add_a_and_b(2, 3);
}
int add_a_and_b(int a, int b) {
return a + b;
}

编译后

image.png

main的指令解释：


push 3	Push the second argument '3' into the stack	把参数3写入栈; ESP
push 2	Push the first argument '2' into the stack	把参数2写入栈
call _add_a_and_b	Call the _add_a_and_b function. This will put the address of the next instruction (add) into the stack, then it will jump into the _add_a_and_b function by putting the address of the first instruction in the _add_a_and_b label (push %ebx) into the EIP register	调用_add_a_and_b函数，会把下一条指令（add）的地址写进栈里，然后通过把_add_a_and_b标签里的第一条指令地址(push %ebx)写入EIP寄存器，来跳转到_add_a_and_b这个函数。
add %esp, 8	Add 8 to the esp, which effectively pops out the two values we just pushed into it	esp寄存器加8个字节，即把push两个value值的空间pop释放掉
ret	Return to the parent function....	返回上一级父层函数。

stack存储状态

For the purposes of this exercise, we’re going to assume that the _main function is located in memory at the address 0xFFFF0000, and that each instructoin is 4 bytes long (the size of each instruction can vary depending on the instruction and on its operands).
So you can see, we first pushed 3 into the stack, %esp was lowered, then we pushed 2 into the stack, %esp was lowered, then we did a ‘call _add_a_and_b’, which stored the address of the next instruction (4 instructions into the main, so ‘_main+16’) into the stack and esp was lowered, then we pushed %ebx, which I assumed here contained a value of 0, and the %esp was lowered again. If we now wanted to access the first argument to the function (2), we need to access %esp+8, which will let us skip the saved %ebx and the ‘Return address’ that are in the stack (since we’re working with 32 bits, each value is 4 bytes). And in order to access the second argument (3), we need to access %esp+12.

翻译如下：

假定main函数在内存里起始地址是0xFFFF0000，每个指令占4个字节（指令大小可以根据指令和操作数变化）

push 3入栈后，ESP指针位置降低（减4）；
然后push 2入栈，ESP又降低（减4）。

接着call _add_a_and_b函数。
所做的操作是：存储main函数里call的下一条指令地址到栈里。
因为call的下一条指令是“add %esp, 8 ”，所以是保存"add %esp, 8"这条指令的地址到stack中，而它在main里是第四条指令，所以是main函数地址+16字节（按一条指令占4字节算）。