我试图通过函数指针表调用一些C++函数,该表作为C符号从共享对象导出。代码实际上正在工作,但是Clang的未定义行为清除器(= UBSan)看到我发出的调用是非法的,如下所示:
==11410==WARNING: Trying to symbolize code, but external symbolizer is not initialized!
path/to/HelloWorld.cpp:25:13: runtime error: call to function (unknown) through pointer to incorrect function type 'foo::CBar &(*)()'
(./libFoo.so+0x20af0): note: (unknown) defined here由于Clang的未定义行为清除器,间接调用函数是合法的,该函数通过函数指针返回C++标准类对象的引用,但对于用户定义的类是非法的。有人能告诉我它怎么了吗?
我一直试图用Clang-llvm3.4-1ubuntu3和CMake 2.8.12.2在Ubuntu14.04上构建这个项目。若要再现此现象,请将以下5文件放在同一个目录中并调用build.sh。它将创建一个makefile并构建项目,并运行可执行文件。
Foo.h
#ifndef FOO_H
#define FOO_H
#include <string>
//
#define EXPORT __attribute__ ((visibility ("default")))
namespace foo {
class CBar
{
// empty
};
class CFoo
{
public:
static CBar& GetUdClass();
static std::string& GetStdString();
};
// function pointer table.
typedef struct
{
CBar& (*GetUdClass)();
std::string& (*GetStdString)();
} fptr_t;
//! function pointer table which is exported.
extern "C" EXPORT const fptr_t FptrInFoo;
}
#endifFoo.cpp
#include "Foo.h"
#include <iostream>
using namespace std;
namespace foo
{
// returns reference of a static user-defined class object.
CBar& CFoo::GetUdClass()
{
cout << "CFoo::GetUdClass" << endl;
return *(new CBar);
}
// returns reference of a static C++ standard class object.
std::string& CFoo::GetStdString()
{
cout << "CFoo::GetStdString" << endl;
return *(new string("Hello"));
}
// function pointer table which is to be dynamically loaded.
const fptr_t FptrInFoo = {
CFoo::GetUdClass,
CFoo::GetStdString,
};
}HelloWorld.cpp
#include <iostream>
#include <string>
#include <dirent.h>
#include <dlfcn.h>
#include "Foo.h"
using namespace std;
using namespace foo;
int main()
{
// Retrieve a shared object.
const string LibName("./libFoo.so");
void *pLibHandle = dlopen(LibName.c_str(), RTLD_LAZY);
if (pLibHandle != 0) {
cout << endl;
cout << "Info: " << LibName << " found at " << pLibHandle << endl;
// Try to bind a function pointer table:
const string SymName("FptrInFoo");
const fptr_t *DynLoadedFptr = static_cast<const fptr_t *>(dlsym(pLibHandle, SymName.c_str()));
if (DynLoadedFptr != 0) {
cout << "Info: " << SymName << " found at " << DynLoadedFptr << endl;
cout << endl;
// Do something with the functions in the function table pointer.
DynLoadedFptr->GetUdClass(); // Q1. Why Clang UBSan find this is illegal??
DynLoadedFptr->GetStdString(); // Q2. And why is this legal??
} else {
cout << "Warning: Not found symbol" << endl;
cout << dlerror() << endl;
}
} else {
cout << "Warning: Not found library" << endl;
cout << dlerror() << endl;
}
cout << endl;
return 0;
}CMakeLists.txt
project (test)
if(COMMAND cmake_policy)
cmake_policy(SET CMP0003 NEW)
endif(COMMAND cmake_policy)
set(CMAKE_SHARED_LINKER_FLAGS "${CMAKE_SHARED_LINKER_FLAGS} -Wl,-rpath,$ORIGIN")
add_library(Foo SHARED Foo.cpp)
add_executable(HelloWorld HelloWorld.cpp)
target_link_libraries (HelloWorld dl)build.sh
#!/bin/bash
# 1. create a build directory.
if [ -d _build ]; then
rm -rf _build
fi
mkdir _build
cd _build
# 2. generate a makefile.
CC=clang CXX=clang++ CXXFLAGS="-fvisibility=hidden -fsanitize=undefined -O0 -g3" cmake ..
# 3. build.
make
# 4. and run the executable.
./HelloWorld我一直试图找到一个线索来挖掘这个问题,并意识到这个问题被消毒液(-fsanitize=function)的“功能”选项捕捉到了,但它并没有那么多的文档。我希望你们能给我一个合理的解释,这样的运行时错误信息,似乎来自另一个星球。谢谢。
在输出中Clang指出的“未知”是什么?
下面是addr2line的输出,用于检查消毒液的“未知”内容:
$ addr2line -Cfe _build/libFoo.so 0x20af0
foo::CFoo::GetUdClass()
path/to/Foo.cpp:12嗯,看起来就像我想给我打电话的那个函数。你能猜到Clang看上去有什么不同吗?
发布于 2015-01-19 09:16:03
CBar的typeinfo需要对函数的类型具有默认的可见性,Clang在可执行文件和动态库中认为相同;将Foo.h更改为:
class EXPORT CBar
{
...
}https://stackoverflow.com/questions/27976687
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