比较 std::endl 的地址
我正在检查一段现有代码，发现使用 Visual C++ 9 和 MinGW 编译时它的行为有所不同：
inline LogMsg& LogMsg::operator<<(std::ostream& (*p_manip)(std::ostream&) )
{
if ( p_manip == static_cast< std::ostream& (*)(std::ostream&) > ( &std::endl<char, std::char_traits<char> >) )
{
msg(m_output.str());
m_output.str( "" );
}
else
{
(*p_manip) (m_output); // or // output << p_manip;
}
return *this;
}
顾名思义，这是一个日志类，它重载 operator<<() 来从流中剥离 endls。
我发现了为什么它的行为不同：测试 p_manip == static_cast... 在 MinGW 中成功，而在 Visual C++ 9 中失败。MinGW
- “忽略”强制转换并返回
的真实地址std::endl; - Visual C++ 9 实际上将指针转换为 endl 并返回不同的地址。
我将测试更改为 if ( p_manip == std::endl ) ，现在它的行为符合预期。
我的问题是：如此复杂（而且事实上是错误的）测试背后的基本原理是什么？
为了完整性：
class LogStream
{
public:
LogStream() {}
protected:
std::ostringstream m_output;
};
class LogMsg : public LogStream
{
friend LogMsg& msg() ;
static LogMsg s_stream;
public:
LogMsg() {}
template <typename T>
inline LogMsg& operator<<(T p_data);
inline LogMsg& operator<<(std::ostream& (*p_manip)(std::ostream&) );
};
I am inspecting a piece of existing code and found out it behaves differently when compiled with Visual C++ 9 and MinGW:
inline LogMsg& LogMsg::operator<<(std::ostream& (*p_manip)(std::ostream&) )
{
if ( p_manip == static_cast< std::ostream& (*)(std::ostream&) > ( &std::endl<char, std::char_traits<char> >) )
{
msg(m_output.str());
m_output.str( "" );
}
else
{
(*p_manip) (m_output); // or // output << p_manip;
}
return *this;
}
As the name suggests, this is a log class and it overloads operator<<() to strip endls from the stream.
I found out why it behaves differently: the test p_manip == static_cast... succeeds with MinGW while it fails with Visual C++ 9.
- MinGW "ignores" the cast and returns the real address of
std::endl; - Visual C++ 9 actually casts the pointer-to-endl and returns a different address.
I changed the test to if ( p_manip == std::endl ) and now it behaves as expected.
My question is: what is the rationale behind such a complicated (and, as a matter of fact, wrong) test?
For the sake of completness:
class LogStream
{
public:
LogStream() {}
protected:
std::ostringstream m_output;
};
class LogMsg : public LogStream
{
friend LogMsg& msg() ;
static LogMsg s_stream;
public:
LogMsg() {}
template <typename T>
inline LogMsg& operator<<(T p_data);
inline LogMsg& operator<<(std::ostream& (*p_manip)(std::ostream&) );
};
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据猜测，我想说原作者没有意识到它们是兼容的类型，并且按照规范进行了转换（编译器没有要求他这样做）。
At a guess, I'd say the original author didn't realise they were compatible types, and did the conversion on spec (without the compiler requiring him to).
有关信息：
语句
if ( p_manip == std::endl )无法在原始编译器（gcc 3.4.5，最初开发代码的编译器）上编译。这意味着测试没有错误，正如我在问题中所说的那样。
For information:
The statement
if ( p_manip == std::endl )does not compile on the original compiler (gcc 3.4.5, the compiler on which the code was originally developed).That means the test was not wrong as I stated in my question.