C ++ 17中的std :: make_shared()更改 [英] std::make_shared() change in C++17

问题描述

在 cppref 中，以下内容一直适用到C ++ 17:

In cppref, the following holds until C++17:

代码(例如f(std::shared_ptr<int>(new int(42)), g()))可能会导致 如果g在new int(42)之后被调用并引发 例外，而f(std::make_shared<int>(42), g())是安全的，因为 两个函数调用永远不会交错.

code such as f(std::shared_ptr<int>(new int(42)), g()) can cause a memory leak if g gets called after new int(42) and throws an exception, while f(std::make_shared<int>(42), g()) is safe, since two function calls are never interleaved.

我想知道C ++ 17中引入的哪些更改使其不再适用.

I'm wondering which change introduced in C++17 renders this no longer applicable.

推荐答案

函数参数的求值顺序由

The evaluation order of function arguments are changed by P0400R0.

在进行更改之前，函数自变量的求值相对于彼此未排序.这意味着g()的评估可能会插入std::shared_ptr<int>(new int(42))的评估中，这会导致您引用的上下文中所述的情况.

Before the change, evaluation of function arguments are unsequenced relative to one another. This means evaluation of g() may be inserted into the evaluation of std::shared_ptr<int>(new int(42)), which causes the situation described in your quoted context.

更改后，函数参数的评估不确定地进行顺序排列，没有交织，这意味着std::shared_ptr<int>(new int(42))的所有副作用都发生在g()的副作用之前或之后.现在考虑g()可能抛出的情况.

After the change, evaluation of function arguments are indeterminately sequenced with no interleaving, which means all side effects of std::shared_ptr<int>(new int(42)) take place either before or after those of g(). Now consider the case where g() may throw.

• 如果std::shared_ptr<int>(new int(42))的所有副作用都早于g()的副作用，则分配的内存将由std::shared_ptr<int>的析构函数释放.

• If all side effects of std::shared_ptr<int>(new int(42)) take place before those of g(), the memory allocated will be deallocated by the destructor of std::shared_ptr<int>.

如果std::shared_ptr<int>(new int(42))的所有副作用都发生在g()的副作用之后，则甚至没有内存分配.

If all side effects of std::shared_ptr<int>(new int(42)) take place after those of g(), there is even no memory allocation.

无论哪种情况，都不会再次发生内存泄漏.

In either case, there is no memory leak again anyway.

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