Debuggable replacement for make_shared()

问题

Using gcc 4.6.2, make_shared() gives a useless backtrace (apparently due to some rethrow) if a constructor throws an exception. I'm using make_shared() to save a bit of typing, but this is show stopper. I've created a substitute make_shrd() that allows a normal backtrace. I'm using gdb 7.3.1.

I'm worried that:

1. The bad backtrace under make_shared() is somehow my own fault
2. My substitute make_shrd() will cause me subtle problems.

Here's a demo:

#include <memory>
#include <stdexcept>

using namespace std;

class foo1
{
public:
        foo1( const string& bar, int x ) :m_bar(bar), m_x(x)
        {
                throw logic_error( "Huh?" );
        }
        string m_bar;
        int m_x;
};

class foo2
{
public:
        foo2( const string& bar, int x ) : m_foo1(bar,x)
        {}

        foo1  m_foo1;
};

// more debuggable substitute for make_shared() ??
template<typename T, typename... Args>
std::shared_ptr<T> make_shrd( Args... args )
{
        return std::shared_ptr<T>( new T(args...));
}

int main()
{
        auto p_foo2 = make_shared<foo2>( "stuff", 5 );          // debug BAD!!
//      auto p_foo2 = make_shrd<foo2>( "stuff", 5 );            // debug OK
//      auto p_foo2 = new foo2( "stuff", 5 );                   // debug OK
//      auto p_foo2 = shared_ptr<foo2>(new foo2( "stuff", 5 )); // debug OK
        return (int)(long int)p_foo2;
}

Compiled with:

g++ -g -std=c++0x -Wall -Wextra main.cpp

Debugged with:

gdb a.out

The make_shared() backtrace is junk that does not show the stack to the point of the exception. All the other options provide a sane backtrace.

Thanks in advance for help and suggestions.

回答1:

Your implementation of make_shrd() looses the ability to allocate just one chunk of memory: std::make_shared() does two things:

1. it avoids duplicating of writing the type (if the type of the allocation and the type of the desired std::shared_ptr<T> are the same rather than the latter being for a base class)
2. it combines allocation of the shared object and the object's descriptor into just one allocation

The main purpose of std::make_shared() is actually the second feature. I haven't looked at the implementation but I suspect that this is also the part which actually causes you problems. Other than that, I don't see any reason why your implementation is any worse once you fix forwarding of arguments:

template<typename T, typename... Args>
std::shared_ptr<T> make_shrd(Args&&... args)
{
    return std::shared_ptr<T>(new T(std::forward<Args>(args)...));
}

来源：https://stackoverflow.com/questions/9233358/debuggable-replacement-for-make-shared