这里发生了什么？我将结果分配给C ++ [英] What is going on here? I assign result to result in C++

问题描述

你能告诉我这里发生了什么吗？以及为什么可行？

Could you tell me what is going on here? And why it is possible?

std::make_unique<int>(1) = std::make_unique<int>(1);

我认为 make_unique 返回了r-值...

I thought that make_unique returned an r-value...

编辑：
您能提供这种构造的一些有用示例吗？

Could you provide some useful example of this construction?

推荐答案

我认为make_unique的返回值是r值...

I thought that return of make_unique is r-value...

是。但是对于像 std :: unique_ptr< ...> 这样的类类型，赋值是对重载成员函数的调用。您可以将右值作为参数传递给函数。更具体地说，是在右值上调用成员函数。

It is. But for class types like std::unique_ptr<...>, the assignment is a call to an overloaded member function. You can pass rvalues as arguments to functions. And more specifically, call member functions on rvalues.

例如，当您检查 std时:: bitset ，您会看到其引用类型实际上是一个类类型。该类型的赋值运算符已重载。它接受 bool ，但实际上执行按位操作来操作 bitset 。

As an example, when you examine std::bitset, you see that its reference type is in fact a class type. The assignment operator for that type is overloaded. It accepts a bool, but in fact performs a bit-wise operation to manipulate a flag in the packed memory of the bitset.

它适用于 unique_ptr 右值，因为它已过载（并且是为其他类类型隐式生成的），而没有任何引用限定符。即

It works for unique_ptr rvalues because it's overloaded (and is implicitly generated for other class types) without any ref-qualifiers. I.e.

unique_ptr& operator=( unique_ptr&& r ) noexcept;

...而不是...

... and not...

unique_ptr& operator=( unique_ptr&& r ) & noexcept;
// Can only be used on lvalues, no overload for rvalues

几乎没有收获。而且，隐式生成的值不可能只用于左值。许多代码（甚至早于C ++ 11的代码）都在rvalues上使用赋值，而没有对操作符进行ref限定。这样会破坏现有的代码库。

Which would have been for very little gain. And it's unlikely that the implicitly generated one would ever be for lvalues only. A lot of code (some that even predates C++11) uses assignment on rvalues, without ref-qualifying the operators. So that would break existing code bases.

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