gdb无法访问内存地址错误 [英] gdb can't access memory address error

问题描述

这里是我的disas码：

  0x0804844d< + 0> ;: push％ebp 
 0x0804844e< + 1>：mov％esp，％ebp 
 0x08048450< + 3> ;:和$ 0xfffffff0，％esp 
 0x08048453< + 6>：sub $ 0x20，％esp 
 0x08048456< ; + 9>：movl $ 0x8048540，（％esp）
 0x0804845d< + 16> ;:呼叫0x8048310< puts @ plt> 
 0x08048462< + 21> ;: lea 0x1c（％esp），％eax 
 0x08048466< + 25> ;: mov％eax，0x4（％esp）
 0x0804846a< + 29> ：movl $ 0x8048555，（％esp）
 0x08048471< + 36> ;: call 0x8048320< scanf @ plt> 
 0x08048476< + 41> ;: mov 0x1c（％esp），％eax 
 0x0804847a< + 45> ;: cmp $ 0x208c，％eax 
 0x0804847f< + 50> ;: jne 0x804848f <主+ 66> 
 0x08048481< + 52> ;: movl $ 0x8048558，（％esp）
 0x08048488< + 59> ;:呼叫0x8048310< puts @ plt> 
 0x0804848d< + 64> ;: jmp 0x804849b< main + 78> 
 => 0x0804848f< + 66>：movl $ 0x8048569，（％esp）
 0x08048496< + 73> ;:呼叫0x8048310< puts @ plt> 
 0x0804849b< + 78>：mov $ 0x0，％eax 
 0x080484a0< + 83> ;:离开
 0x080484a1< + 84> ;: ret 
  

我要研究的是$ 0x208c。当我键入x / xw 0x208c时，它给了我一个错误信息，表示无法访问地址0x208c处的内存。当我输入Info寄存器并查看eax时，它会显示我提供的值。所以基本上这个程序比较两个值，并根据这些值打印出来。问题是，这是来自大学的作业，我没有得到代码。希望你能帮助。

解决方案

当我键入 x / xw 0x208c 它给我回错误说无法访问地址为0x208c的内存




你的程序的反汇编说它做了这样的事情：



  puts（some string）; 
 int i; 
 scanf（％d，& i）; //我不知道实际的格式字符串是什么。 
 //你可以找到x / s 0x8048555 
 if（i == 0x208c）{...} else {...} 
  

换句话说， 0x208c 是一个值（ 8332 ）你的程序已经硬编码了，而且不是指针。因此，GDB完全正确地告诉你，如果你将 0x208c 解释为一个指针，那么这个指针并不指向可读的内存。

我终于想出了使用print语句而不是x / xw

不理解 print 和 examine 命令之间的区别。考虑这个例子：



  int foo = 42; 
 int * pfoo =& foo; 
  

如上所述， print pfoo 会给你 foo 和 x pfoo 的地址会为您提供值存储在该地址（即 foo 的值）。


here is my disas code:

   0x0804844d <+0>:     push   %ebp
   0x0804844e <+1>:     mov    %esp,%ebp
   0x08048450 <+3>:     and    $0xfffffff0,%esp
   0x08048453 <+6>:     sub    $0x20,%esp
   0x08048456 <+9>:     movl   $0x8048540,(%esp)
   0x0804845d <+16>:    call   0x8048310 <puts@plt>
   0x08048462 <+21>:    lea    0x1c(%esp),%eax
   0x08048466 <+25>:    mov    %eax,0x4(%esp)
   0x0804846a <+29>:    movl   $0x8048555,(%esp)
   0x08048471 <+36>:    call   0x8048320 <scanf@plt>
   0x08048476 <+41>:    mov    0x1c(%esp),%eax
   0x0804847a <+45>:    cmp    $0x208c,%eax
   0x0804847f <+50>:    jne    0x804848f <main+66>
   0x08048481 <+52>:    movl   $0x8048558,(%esp)
   0x08048488 <+59>:    call   0x8048310 <puts@plt>
   0x0804848d <+64>:    jmp    0x804849b <main+78>
=> 0x0804848f <+66>:    movl   $0x8048569,(%esp)
   0x08048496 <+73>:    call   0x8048310 <puts@plt>
   0x0804849b <+78>:    mov    $0x0,%eax
   0x080484a0 <+83>:    leave  
   0x080484a1 <+84>:    ret 

what i'm tring to examine is $0x208c. When I type x/xw 0x208c it gives me back error which says Cannot access memory at address 0x208c. When i type Info registers and look at eax it says the value which i provided. So basically this program compares two values and depending on that prints something out.The problem is that this is homework from university and I have not got code. Hope you can help. Thank you.

解决方案

When I type x/xw 0x208c it gives me back error which says Cannot access memory at address 0x208c

The disassembly for your program says that it does something like this:

puts("some string");
int i;
scanf("%d", &i);  // I don't know what the actual format string is.
                  // You can find out with x/s 0x8048555
if (i == 0x208c) { ... } else { ... }

In other words, the 0x208c is a value (8332) that your program has hard-coded in it, and is not a pointer. Therefore, GDB is entirely correct in telling you that if you interpret 0x208c as a pointer, that pointer does not point to readable memory.

i finally figured out to use print statement instead of x/xw

You appear to not understand the difference between print and examine commands. Consider this example:

int foo = 42;
int *pfoo = &foo;

With above, print pfoo will give you the address of foo, and x pfoo will give you the value stored at that address (i.e. the value of foo).

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