“无抛出取消引用”； std :: unique_ptr的 [英] "No-throw dereferencing" of std::unique_ptr

问题描述

我用C ++编写代码，该代码使用 std :: unique_ptr u 处理 std :: string 资源，并且我想取消引用 u ，以便可以将 std :: string 传递给<$的调用c $ c> std :: string 复制构造函数：

I write code in C++ which uses a std::unique_ptr u to handle a std::string resource, and I want to dereference u so that I can pass the std::string to a call of the std::string copy constructor:

std::string* copy = new std::string( /*dereference u here*/ );

我知道 new 或 std :: string 复制构造函数可能会抛出，但这不是我的意思。我只是想知道，取消引用 u 是否已经可以引发异常。我觉得奇怪的是， operator * 被 not 标记为 noexcept 而 std :: unique_ptr 方法 get 实际上被标记为 noexcept 。换句话说：

I know that new or the std::string copy constructor could throw, but this is not my point here. I was just wondering whether dereferencing u could already throw an exception. I find it strange that operator* is not marked noexcept while the std::unique_ptr method get is actually marked noexcept. In other words:

*( u.get() )

整体上为 ，而

*u

不是。这是标准的缺陷吗？我不明白为什么会有区别。有想法吗？

isn't. Is this a flaw in the standard? I don't get why there could be a difference. Any ideas?

推荐答案

unique_ptr :: operator *（）可能涉及调用 operator *（）重载，调用您存储在 unique_ptr 中的类型。请注意，存储在 unique_ptr 中的类型不必是裸指针，可以通过嵌套类型 D :: pointer ，其中 D 是 unique_ptr 的删除程序的类型。这就是为什么函数不是 noexcept 的原因。

unique_ptr::operator*() could involve a call to an operator*() overload for the type you're storing in the unique_ptr. Note that the type stored in a unique_ptr need not be a bare pointer, you can change the type via the nested type D::pointer, where D is the type of the unique_ptr's deleter . This is why the function is not noexcept.

此警告不适用于您的用例，因为您将 std :: string * 重新存储在 unique_ptr 中，而不是将运算符重载的某些类型* 。因此，通话实际上对您来说是 noexcept 。

This caveat doesn't apply to your use case because you're storing an std::string * in the unique_ptr and not some type that overloads operator*. So the call is effectively noexcept for you.

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