make_shared()的可调试替换 [英] Debuggable replacement for make_shared()

问题描述

使用gcc 4.6.2，如果构造函数引发异常，make_shared()会给出无用的回溯(显然是由于重新抛出).我正在使用make_shared()节省一些键入内容，但这是show stopper.我创建了一个替代 make_shrd()，它可以进行正常的回溯.我正在使用gdb 7.3.1.

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.

我担心:

1. make_shared()下不好的回溯是我自己的错
2. 我的替代品make_shrd()会给我带来一些细微的问题.

这是一个演示:

#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;
}

编译:

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

已调试:

gdb a.out

make_shared()回溯是垃圾，无法显示堆栈到异常点.所有其他选项提供了合理的回溯.

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.

预先感谢您的帮助和建议.

Thanks in advance for help and suggestions.

推荐答案

您对 make_shrd()的实现失去了仅分配一块内存的功能: std :: make_shared()做两件事:

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

1. 避免重复编写类型(如果分配的类型和所需的 std :: shared_ptr< T> 的类型相同，则后者是基类的)
2. 它将共享对象和对象描述符的分配合并为一个分配

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

std :: make_shared()的主要目的实际上是第二个功能.我没有看过实现，但是我怀疑这实际上也是导致您出现问题的部分.除此之外，一旦修复了参数转发问题，我看不出您的实现会变得更糟的任何原因:

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)...));
}

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