rja*_*cks 4 c c++ instrumentation gcc
我最近读到了使用GCC的代码生成功能(特别是-finstrument-functions编译器标志)来轻松地将检测添加到我的程序中.我认为这听起来很酷,并且在以前的C++项目中尝试过.经过我的补丁的几次修改后,我发现每当我尝试使用STL容器或使用C++流I/O打印到stdout时,我的程序会立即崩溃并发生段错误.我的第一个想法是要保持std::list
的Event
结构
typedef struct
{
unsigned char event_code;
intptr_t func_addr;
intptr_t caller_addr;
pthread_t thread_id;
timespec ts;
}Event;
list<Event> events;
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程序终止时将写入文件.GDB告诉我,当我尝试添加一个Event
列表时,调用events.push_back(ev)
自己启动了一个检测调用.在考虑了一下之后,这并不令人惊讶并且有意义,所以计划2.
博客中让我参与所有这些混乱的例子没有做任何疯狂的事情,它只是简单地将一个字符串写入文件中fprintf()
.我不认为使用C++的基于流的I/O而不是旧的I/O会有任何损害(f)printf()
,但这种假设被证明是错误的.这一次,GDB报告了标准库中相当正常的下降,而不是几乎无限的死亡螺旋,然后是段错误.
#include <list>
#include <iostream>
#include <stdio.h>
using namespace std;
extern "C" __attribute__ ((no_instrument_function)) void __cyg_profile_func_enter(void*, void*);
list<string> text;
extern "C" void __cyg_profile_func_enter(void* /* unused */, void* /* unused */)
{
// Method 1
text.push_back("NOPE");
// Method 2
cout << "This explodes" << endl;
// Method 3
printf("This works!");
}
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#0 _int_malloc (av=0x7ffff7380720, bytes=29) at malloc.c:3570
#1 0x00007ffff704ca45 in __GI___libc_malloc (bytes=29) at malloc.c:2924
#2 0x00007ffff7652ded in operator new(unsigned long) ()
from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#3 0x00007ffff763ba89 in std::string::_Rep::_S_create(unsigned long, unsigned long, std::allocator<char> const&) () from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#4 0x00007ffff763d495 in char* std::string::_S_construct<char const*>(char const*, char const*, std::allocator<char> const&, std::forward_iterator_tag) () from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#5 0x00007ffff763d5e3 in std::basic_string<char, std::char_traits<char>, std::allocator<char> >::basic_string(char const*, std::allocator<char> const&) () from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#6 0x00000000004028c1 in __cyg_profile_func_enter () at src/instrumentation.cpp:82
#7 0x0000000000402c6f in std::move<std::string&> (__t=...) at /usr/include/c++/4.6/bits/move.h:82
#8 0x0000000000402af5 in std::list<std::string, std::allocator<std::string> >::push_back(std::string&&) (this=0x6055c0, __x=...) at /usr/include/c++/4.6/bits/stl_list.h:993
#9 0x00000000004028d2 in __cyg_profile_func_enter () at src/instrumentation.cpp:82
#10 0x0000000000402c6f in std::move<std::string&> (__t=...) at /usr/include/c++/4.6/bits/move.h:82
#11 0x0000000000402af5 in std::list<std::string, std::allocator<std::string> >::push_back(std::string&&) (this=0x6055c0, __x=...) at /usr/include/c++/4.6/bits/stl_list.h:993
#12 0x00000000004028d2 in __cyg_profile_func_enter () at src/instrumentation.cpp:82
#13 0x0000000000402c6f in std::move<std::string&> (__t=...) at /usr/include/c++/4.6/bits/move.h:82
#14 0x0000000000402af5 in std::list<std::string, std::allocator<std::string> >::push_back(std::string&
...
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#0 0x00007ffff76307d1 in std::ostream::sentry::sentry(std::ostream&) ()
from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#1 0x00007ffff7630ee9 in std::basic_ostream<char, std::char_traits<char> >& std::__ostream_insert<char, std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&, char const*, long) ()
from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#2 0x00007ffff76312ef in std::basic_ostream<char, std::char_traits<char> >& std::operator<< <std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&, char const*) ()
from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#3 0x000000000040251e in __cyg_profile_func_enter () at src/instrumentation.cpp:81
#4 0x000000000040216d in _GLOBAL__sub_I__ZN8GLWindow7attribsE () at src/glwindow.cpp:164
#5 0x0000000000402f2d in __libc_csu_init ()
#6 0x00007ffff6feb700 in __libc_start_main (main=0x402cac <main()>, argc=1, ubp_av=0x7fffffffe268,
init=0x402ed0 <__libc_csu_init>, fini=<optimized out>, rtld_fini=<optimized out>,
stack_end=0x7fffffffe258) at libc-start.c:185
#7 0x0000000000401589 in _start ()
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cout
在instrumentation函数中使用的问题是,__libc_csu_init()
在全局C++对象有机会被构造之前,调用instrumentation函数是运行时初始化的一个非常早期的部分(事实上,我认为__libc_csu_init()
负责启动那些构造函数 - 至少是间接的).
所以cout
还没有机会进行构建,尝试使用它并不能很好地工作......
这可能是您std::List
在修复无限递归后尝试使用时遇到的问题(在Dave S的回答中提到).
如果您在初始化期间愿意丢失一些仪器,您可以执行以下操作:
#include <iostream>
#include <stdio.h>
int initialization_complete = 0;
using namespace std;
extern "C" __attribute__ ((no_instrument_function)) void __cyg_profile_func_enter(void*, void*);
extern "C" void __cyg_profile_func_enter(void* /* unused */, void* /* unused */)
{
if (!initialization_complete) return;
// Method 2
cout << "This explodes" << endl;
// Method 3
printf("This works! ");
}
void foo()
{
cout << "foo()" << endl;
}
int main()
{
initialization_complete = 1;
foo();
}
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