我在记忆中有一个地址,我想找出该记忆地址的权限(r/w/x)。
例如:
char *s = "hello";
在这里, 字符串字串“ hello” 存储在只读存储器中。 当通过 gdb 运行程序时, 是否有可能检查该内存地址( 是否只允许读 ) 的权限?
我在记忆中有一个地址,我想找出该记忆地址的权限(r/w/x)。
例如:
char *s = "hello";
在这里, 字符串字串“ hello” 存储在只读存储器中。 当通过 gdb 运行程序时, 是否有可能检查该内存地址( 是否只允许读 ) 的权限?
您可以首先找到 s
指向的地方 :
(gdb) print s
$6 = 0x400dbc "foo"
然后发现其中的一节:
(gdb) maintenance info sections
Exec file:
`/home/mfukar/tmp , file type elf64-x86-64.
...sections...
0x00400db8->0x00400dfb at 0x00000db8: .rodata ALLOC LOAD READONLY DATA HAS_CONTENTS
...more sections...
查找 readly
旗帜。
或者,请查看 /proc/PID/maps
,其中 p > pID
是您重新调试和可以使用 info proc
获得的过程的中间点。
For example, let s say I want to find a particular word or number in a file. The contents are in sorted order (obviously). Since I want to run a binary search on the file, it seems like a real waste ...
最好、最小、最快、开放的来源、C/C++ 3d 提供方(在3ds max模型的支持下),而不是通用公平市价,
Given a 10 digit Telephone Number, we have to print all possible strings created from that. The mapping of the numbers is the one as exactly on a phone s keypad. i.e. for 1,0-> No Letter for 2->...
I m trying to find the source of a bug I have found in an open-source application. I have managed to get a build up and running on my Windows machine, but I m having trouble finding the spot in the ...
I wrote below code to readin line by line from stdin ex. city=Boston;city=New York;city=Chicago and then split each line by ; delimiter and print each record. Then in yet another loop I try to ...
I was wondering if there were any good free graphics libraries for C that are easy to use? It s for plotting 2d and 3d graphs and then saving to a file. It s on a Linux system and there s no gnuplot ...
Is there anything other than DDD that will draw diagrams of my data structures like DDD does that runs on Linux? ddd is okay and runs, just kind of has an old klunky feeling to it, just wanted to ...
Please note that this is not homework and i did search before starting this new thread. I got Store an int in a char array? I was looking for an answer but didn t get any satisfactory answer in the ...