使用GDB并检查数据的内存布局 [英] Using GDB and checking the memory layout of Data
问题描述
假设我们有一个简单的C ++代码,如下所示:
Assume we have a simple C++ code as the following:
#include <iostream>
int main(){
int a = 5;
}
由于每个内存位置都是8 bits
,整数是32 bits
,所以我假设a
的内存结构是这样的:
Since each memory location is 8 bits
and an integer is 32 bits
I assume the memory structure for a
would be like this:
0xa 0xb 0xc 0xd
00000000 00000000 00000000 00000101
其中0xa,0xb,0xc,0xd是示例内存地址.
where 0xa,0xb,0xc,0xd are sample memory addresses.
1)是&a
指向0xa
还是0xd
?
2)如果我使用GDB并使用x
来获取真实的内存地址,则会得到以下信息:
2) if I use GDB and and use x
to get real memory addresses I get the following:
(gdb) p a
$7 = 5
(gdb) p &a
$8 = (int *) 0x7ffeefbffac8
(gdb) x/bt 0x7ffeefbffac8
0x7ffeefbffac8: 00000101
(gdb) x/bt 0x7ffeefbffac8-1
0x7ffeefbffac7: 00000000
(gdb) x/bt 0x7ffeefbffac8-2
0x7ffeefbffac6: 00000000
(gdb) x/bt 0x7ffeefbffac8-3
0x7ffeefbffac5: 01111111
(gdb)
为什么0x7ffeefbffac8-3
用01111111
而不是00000000
填充?这个地址不等于我们的示例内存地址中的0xa
吗?
why is 0x7ffeefbffac8-3
populated with 01111111
and not 00000000
? ins't this address equal to 0xa
in our sample memory address?
推荐答案
在小端机上,& a指向内存的最低有效字节.也就是说,如果&a == 0x7ffeefbffac8
,则a
驻留在字节中
On a little-endian machine, &a points to the least significant byte of memory. That is, if &a == 0x7ffeefbffac8
, then a
resides in bytes
0x7ffeefbffac8: 101 << least significant byte
0x7ffeefbffac9: 000
0x7ffeefbffaca: 000
0x7ffeefbffacb: 000 << most significant byte.
最好通过分配例如0x0a090705
到a
,然后:
This is best observed by assigning e.g. 0x0a090705
to a
, and then:
Temporary breakpoint 1, main (argc=3, argv=0x7fffffffdc68) at t.c:2
2 int a = 0x0a090705;
(gdb) n
3 return 0;
(gdb) p &a
$1 = (int *) 0x7fffffffdb7c
检查从& a开始的4个字节:
Examine 4 bytes starting from &a:
(gdb) x/4bt 0x7fffffffdb7c
0x7fffffffdb7c: 00000101 00000111 00001001 00001010
或者等效地,一次执行一个字节:
Or, equivalently, do so one byte at a time:
(gdb) x/bt 0x7fffffffdb7c
0x7fffffffdb7c: 00000101
(gdb) x/bt 0x7fffffffdb7c+1
0x7fffffffdb7d: 00000111
(gdb) x/bt 0x7fffffffdb7c+2
0x7fffffffdb7e: 00001001
(gdb) x/bt 0x7fffffffdb7c+3
0x7fffffffdb7f: 00001010
为什么
0x7ffeefbffac8-3
用01111111
而不是00000000
填充?
因为您走错了方向:&a-3
根本不是a
的一部分(这是其他东西的一部分,或者可能是未初始化的随机垃圾).
Because you are going in the wrong direction: &a-3
isn't part of a
at all (it's part of something else, or possibly uninitialized random garbage).
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