Question

我想了解一个类'ostream运算符“存在于对象文件中的位置。在下面的代码示例中，我设计了C ++类DeltaTimer，它是朋友的ostream运算符。下面是“nm DeltaTimer.o”的输出。在此输出中对“ostream”进行Grepping会显示符号类型“U”，即未定义。但代码编译（main.cpp包含完整性）和链接成功，并按预期执行，这意味着没有任何未定义。但是我不明白ostream运算符在“nm”的输出中解析了什么，并且如果有人可以提供建议，我将不胜感激。谢谢。

DeltaTimer.h：

template<>
struct traits<int> {
   static double func()
   {
      return f();
   }
};

template<>
struct traits<double> {
   static double func()
   {
      return g();
   }
};

DeltaTimer.cpp：

#ifndef DeltaTimer_h
#define DeltaTimer_h

#include <iostream>

class DeltaTimer
{
public:
    DeltaTimer();
    virtual ~DeltaTimer();
    void Foo();

private:
    int m_foo;

friend std::ostream& operator<<(std::ostream& os, const DeltaTimer& rDeltaTimer);
};

#endif // DeltaTimer_h

main.cpp中：

#include "DeltaTimer.h"
#include <iostream>

DeltaTimer::DeltaTimer() : m_foo(5)
{
    std::cout << "DeltaTimer ctor" << std::endl;
}

DeltaTimer::~DeltaTimer()
{
    std::cout << "DeltaTimer dtor" << std::endl;
}

void DeltaTimer::Foo()
{
    std::cout << "DeltaTimer.m_foo == " << m_foo << std::endl;
}

std::ostream& operator<<(std::ostream& os, const DeltaTimer& rDeltaTimer)
{
    os << "DeltaTimer[m_foo(" << rDeltaTimer.m_foo << ")]" << std::endl;
}

// This function is just a sanity-check for the output of "nm"
void Bar()
{
    std::cout << "void Bar()" << std::endl;
}

输出“nm”：

#include "DeltaTimer.h"
#include <iostream>

int main()
{
    DeltaTimer deltaTimer;

    std::cout << deltaTimer << std::endl;
}

Answer 1

这是因为nm显示了错误的名字。要打印人类可读的，您需要使用--demangle参数：

$ nm --demangle DeltaTimer.o 
                 U __cxa_atexit
                 U __dso_handle
00000000000001a4 t _GLOBAL__sub_I__ZN10DeltaTimerC2Ev
                 U __gxx_personality_v0
0000000000000145 T Bar()
0000000000000167 t __static_initialization_and_destruction_0(int, int)
                 U operator delete(void*)
00000000000000f1 T operator<<(std::ostream&, DeltaTimer const&) // <== here's your guy!
00000000000000b0 T DeltaTimer::Foo()
0000000000000000 T DeltaTimer::DeltaTimer()
0000000000000000 T DeltaTimer::DeltaTimer()
000000000000008a T DeltaTimer::~DeltaTimer()
0000000000000040 T DeltaTimer::~DeltaTimer()
0000000000000040 T DeltaTimer::~DeltaTimer()
                 U std::ostream::operator<<(int)
                 U std::ostream::operator<<(std::ostream& (*)(std::ostream&))
                 U std::ios_base::Init::Init()
                 U std::ios_base::Init::~Init()
                 U std::cout
                 U std::basic_ostream<char, std::char_traits<char> >& std::endl<char, std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&)
0000000000000000 r std::piecewise_construct
0000000000000000 b std::__ioinit
                 U std::basic_ostream<char, std::char_traits<char> >& std::operator<< <std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&, char const*)
0000000000000000 V typeinfo for DeltaTimer
0000000000000000 V typeinfo name for DeltaTimer
0000000000000000 V vtable for DeltaTimer
                 U vtable for __cxxabiv1::__class_type_info

或者只是使用c++filt。具体做法是：

$ echo _ZlsRSoRK10DeltaTimer | c++filt
operator<<(std::basic_ostream<char, std::char_traits<char> >&, DeltaTimer const&)

对象文件中的ostream运算符会发生什么？

1 个答案: