我的项目中的所有异常都有一个基类。它是这样实现的:
异常.hpp
class Exception : public std::exception
{
public:
ELS_EXPORT_SYMBOL Exception(void) throw();
ELS_EXPORT_SYMBOL explicit Exception(const std::string& what) throw();
ELS_EXPORT_SYMBOL Exception(const char* format, ...) throw()
ELS_PRINTF_FUNC(2, 3);
ELS_EXPORT_SYMBOL Exception(const Exception& other) throw();
ELS_EXPORT_SYMBOL Exception& operator =(const Exception& other)
throw();
ELS_EXPORT_SYMBOL virtual ~Exception(void) throw();
ELS_EXPORT_SYMBOL virtual const char* what(void) const throw();
protected:
ELS_EXPORT_SYMBOL void _M_setWhat(const char* format, ::va_list va)
throw();
private:
std::string _M_what;
};
#define ELS_EXC_VA_SET_WHAT(FORMAT) \
do \
{ \
::va_list va; \
::va_start(va, FORMAT); \
this->_M_setWhat(FORMAT, va); \
::va_end(va); \
} \
while (false)
异常.cpp
Exception::Exception(void) throw()
: std::exception(),
_M_what("Exception")
{
}
Exception::Exception(const std::string& what) throw()
: std::exception(),
_M_what(what.empty() ? "Exception" : what)
{
}
Exception::Exception(const char* format, ...) throw()
: std::exception(),
_M_what()
{
ELS_EXC_VA_SET_WHAT(format);
}
Exception::Exception(const Exception& other) throw()
: std::exception(other),
_M_what(other._M_what)
{
}
Exception& Exception::operator =(const Exception& other) throw()
{
std::exception::operator =(other);
this->_M_what = other._M_what;
return *this;
}
Exception::~Exception(void) throw()
{
}
const char* Exception::what(void) const throw()
{
return this->_M_what.c_str();
}
void Exception::_M_setWhat(const char* format, ::va_list va) throw()
{
static const size_t BUFSIZE = 512;
char buf[BUFSIZE];
::memset(buf, 0, BUFSIZE);
::vsnprintf(buf, BUFSIZE, format, va);
this->_M_what = std::string(buf);
}
这些文件是使用以下选项编译的共享库的一部分:
-Wall -fPIC -O2 -D_GNU_SOURCE -fvisibility=hidden -rdynamic -Wl,-E
库本身编译没有任何问题,但是当我尝试编译二进制文件并将其链接到它时,我收到以下错误:
Main.cpp:(.text+0x297): undefined reference to `typeinfo for els::except::Exception'
我用谷歌搜索了一下,常见的答案是要么实现所有虚拟方法,要么使它们成为纯方法。我在这里看不到这个问题。我究竟做错了什么?