如果隐式生成的赋值运算符是& ref-qualified? [英] Should implicitly generated assignment operators be & ref-qualified?
问题描述
以下代码在gcc 4.8.1上没有问题地编译:
The following code compiles without problem on gcc 4.8.1:
#include <utility>
struct foo
{
};
int main()
{
foo bar;
foo() = bar;
foo() = std::move( bar );
}
似乎隐式生成的赋值运算符 foo
不是&
ref-qualified,因此可以在右值上调用。根据标准这是正确的吗?如果是,原因是不是要求隐式生成的赋值运算符&
ref-qualified?
It seems the implicitly generated assignment operators for foo
are not &
ref-qualified and so can be invoked on rvalues. Is this correct according to the standard? If so, what reason is there for not requiring implicitly generated assignment operators to be &
ref-qualified?
为什么标准不需要生成以下内容?
Why doesn't the standard require the following to be generated?
struct foo
{
foo & operator=( foo const & ) &;
foo & operator=( foo && ) &;
};
推荐答案
一个右值。引用 作业的参考限定符标准库中的运算符:
Well, there are certain legitimate use cases for assigning to an rvalue. To quote from Ref-qualifiers for assignment operators in the Standard Library:
只有几种非常具体的类型,
支持赋值到右值。特别地,用作
代理的类型,例如向量< bool> ::引用,以及其赋值
运算符是const限定的类型(例如,slice_array)。
There are only a few very specific types for which it makes sense to support assigning to an rvalue. In particular, types that serve as a proxy, e.g., vector<bool>::reference, and types whose assignment operators are const-qualified (e.g., slice_array).
C ++标准委员会显然认为默认赋值不应该有一个隐式的ref限定符, / em>。事实上,如果突然间所有隐式声明的赋值运算符都不能使用右值,那么可能存在现有的代码会停止工作。
The C++ standard committee obviously felt that default assignment should not have an implicit ref qualifier - rather it should be explicitly declared. Indeed, there may be existing code which would stop working if all of a sudden all implicitly declared assignment operators didn't work with rvalues.
授予,有点难提出了一个例子,我们想要一个隐式声明的赋值运算符使用右值,但C ++标准委员会可能不想采取这些类型的机会,当涉及到保持向后兼容性。这样的代码:
Granted, it's a bit hard to contrive an example where we want an implicitly declared assignment operator to work with rvalues, but the C++ standard committee likely doesn't want to take these kind of chances when it comes to preserving backwards compatibility. Code like this:
int foo_counter = 0;
struct Foo
{
Foo()
{
++foo_counter;
}
~Foo()
{
--foo_counter;
}
};
int main()
{
Foo() = Foo();
}
...不再工作了。在一天结束时,标准委员会希望确保以前有效的C ++(无论多么愚蠢或诡诈)在C ++ 11中继续工作。
...wouldn't work anymore. And at the end of the day, the standards committee wants to make sure that previously valid C++ (no matter how stupid or contrived) continues to work in C++11.
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